LA NMO 2005 A FOST FACUT  UN ANUNT CUM CA LA  JUNIORI  (NUMAI  LA CLS A7A) SE FORMEAZA UN LOT LA CAMPULUNG NUMIT "LOTUL SPERANTEI" ALCATUIT DIN PRIMII 20 CLASATI LA NATIONALA LA CLS A7A. ACESTIA VOR DA SIMULTAN BARAJELE CU CEI DE LA BUCURESTI.REALIZAREA UNUI ASTFEL DE LOT ESTE O PREMIERA SI AS DORI MAI MULTE DETALII SAU EVENTUAL O CONFIRMARE A ACESTUI ANUNT.

Inainte de confirmare te rog sa nu mai folosesti in tot textul literele capitale. Revenind la subiect, pot sa confirm ca Domnul Branzei a distribuit o hartie prin care se propunea crearea acestui "lot al sperantei" cu elevi de clasa a 7-a, dintre cei care nu au intrat in lotul olimpic de juniori, care sa fie pregatit la Campul Lung-Moldovenesc (probabil de catre Prof. Branzei, Prof. Balauca, Prof. Piticari, si alti profesori de talie ce rezideaza in zona nordica a Moldovei).

Ideea era ca scolile sa suporte chetuielile de cazare si masa, precum si transport pentru fiecare elev in parte. Din cate imi amintesc acestea se ridicau la cca 10 lei noi / zi cazare + 15 lei / zi masa pentru un elev, deci pentru un "sejur" de 10 zile, cca 250 de lei noi / elev. Fara acordul scolilor evident ca acest lucru nu se poate realiza.

Personal cred ca este o idee excelenta. Eu am beneficiat de un astfel de program in 1998, cand lotul de juniori (primul creat) s-a tinut pentru 1 luna intreaga la Iasi, Bucium, si aveam in fiecare zi cursuri cu Dnul Branzei, Dnul Enescu, si multi alti profesori care au trecut si si-au pus o amprenta asupra pregatirii noastre.

[quote="christy-metmat"]Daca sunteti dintre voi calificati in lotul acesta mi-ar placea sa luam legatura... :)[/quote]
   Teoretic ar trebui si eu sa fac parte din acest lot, dar nici eu nu stiu prea multe despre el, daca va fi oare cu adevarat, sau va ramane doar la stadiul de tentativa :?
   Daca cineva afla informatii in plus as fi  :)  sa mi le impartaseasca si mie...
          Hello, va las ca e tarziu-oricum maine revin!

Am scris mai sus cam care sunt informatiile disponibile in acest moment. O sa incerc sa iau legatura cu prof. Branzei sa vad daca mai sunt detalii. Problema in principal este de ordin financiar: scolile la care studiati voi cei de a-7-a trebuie sa plateasca pentru cazarea si masa voastra la Campu-Lung Moldovenesc pentru o perioada (pentru inceput) de cca 2 saptamani (faceti voi aritmetica singuri cat vine asta :) ).

O sa incerc sa iau legatura cu prof. Branzei si sa vedem ce se intampla exact, dar mi-e teama ca mai mult de ce am scris si eu mai sus nu o sa fie deocamdata ...  :?

[quote="christy-metmat"]felixmann, de unde esti si al catelea ai fost la NMO?am mai primit niste informatii de la Dragos Constantinescu. Pana una alta  pastram legatura nu?
 sper  ca intr-adevar sa se poata realiza acest lot[/quote]
Si eu ma numar printre cei interesati de acest subiect  ;)

:(  :(  :(
Am o veste proasta...
Noi nu o sa mai mergem la pregatirea de la Campulung-Moldovenesc in per 16-24 aprilie ,deoarece nu au fost facute ultimele aranjamente de catre minister,dar vom merge la urmatoarele etape de pregatire si la barajele de la Bucuresti{care sunt mai importante, dupa parerea mea}.
Sper sa mai vorbim prin intermediul  net-ului pana atunci.

[quote] vom merge la urmatoarele etape de pregatire si la barajele de la Bucuresti[/quote]

Spune-mi te rog, ce inseamna concret "urmatoarele etape de pregatire si la barajele de la Bucuresti"? In ce perioada se vor intampla?

Multumesc.

[quote="Valentin Vornicu"]Poate nu am inteles eu bine, dar pe hartia pe care am avut-o scria clar "primi 20 care [b]NU[/b] au intrat in lotul olimpic de juniori de la clasa a VII-a". Deci din start daca ai intrat in lot, nu intrai in vizor :)[/quote]

Asa am inteles si eu, anume ca acest "Lotul sperantei" se refera la primii 20 care NU au intrat in lotul olimpic de juniori, iar eu ma incadrez la aceasta categorie. Am mai inteles ca a fost o propunere ca "Lotul sperantei" sa participe la un fel de tabara de pregatire la Campulung-Moldovenesc, dar dupa cum reiese si din discutiile de mai sus, propunerea nu s-a materializat.
De aceea, cand am vazut cele scrise de christy-metmat, cum ca "nu o sa mai mergem la pregatirea de la Campulung-Moldovenesc [...] dar vom merge la urmatoarele etape de pregatire si la barajele de la Bucuresti", am vrut sa aflu mai multe.  Presupun ca christy-metmat se referea la lotul sperantei. Sau am inteles eu gresit?

Multumesc.

Cred ca christy-metmat este in lotul olimpic si nu stia exact de clauza de mai sus.

[quote="christy-metmat"]Astfel ar putea fi o posibilitate ca sa pot da si eu urmatoarele baraje.[/quote]Daca nu ai intrat in lot nu ai cum sa dai barajele. Asta e scopul lotului largit, si al barajelor ce se tin la nationala.

Inveti mai mult si la anul faci mai bine. Asta e.

[quote="christy-metmat"]Nu stiu cine credeti voi ca sunt-dar eu nu am intrat in lotul olimpic ,de fapt eu am fost a doua sub  ultimul care s-a dus la baraje. :(  :?
Acesta ar fi si unul din motivele pentru care vreau sa aflu mai multe informatii despre acest lot.Astfel ar putea fi o posibilitate ca sa pot da si eu urmatoarele baraje.[/quote]
Eu am avut 24p la NMO - clasa a VII-a, dar nu am intrat in lotul olimpic de juniori, deci indeplinesc criteriile pt "Lotul sperantei"  ;)
Asa ca in continuare ma intereseaza acest subiect.

Se pare ca totusi initiativa cu "Lotul sperantei" nu a fost abandonata  :lol:
Se va organiza o tabara de pregatire la Campulung Moldovenesc, in perioada 1-20 iulie, pentru aproximativ 25 elevi de clasa a VII-a care au obtinut rezultate bune la faza nationala, dar nu s-au calificat pentru balcaniada de juniori.
Programul de pregatire va cuprinde in principiu 6 ore de mate/zi 5zile/saptamana, pe langa alte activitati cultural-sportive.
Cade in sarcina scolilor respective sa se ocupe de finantarea elevilor (250.000 lei/zi/elev => 5 milioane lei in total/elev).
Deocamdata nu stiu mai mult  ;)

[quote="xtreme_boy90"]SUNT SI EU IN LOTUL SPERANTEI .AM LUAT AL 26-LEA LA NATIONALA.NU STIU EXACT DACA FACEM CURSURI VINERI,SAU E DOAR SOSIREA PARTICIPANTILOR.DACA AFLII SPUNE-MI SI MIE . :lol:[/quote]
Ceao, xtreme_boy90. Ma bucur ca vii in tabara.
Din pacate nu am o informatie oficiala legata de data de 1 iulie. Tot ce pot sa spun este ca am luat legatura cu dl. prof. Cerbu de la Campulung, care raspunde de partea de cazare. Dumnealui insa nu stia care este programul cursurilor. Din cate am inteles, ideal ar fi sa ajungem in 30 seara, dar credea ca nu e o nenorocire daca cineva nu reuseste sa ajunga decat in 1, mai ales cu greva asta de la CFR.
Eu sunt din Timisoara si pot sa spun ca nici un accelerat Timisoara-Iasi nu circula. Mai mult, nici un accelerat care opreste in Campulung nu circula. Am fost la agentie si mi s-a spus sa vin in 29, asa ca pana atunci nu stiu cand si cum voi putea ajunge  :(

[quote="Valentin Vornicu"]Deci pana la urma se tine ceva? :)[/quote]
Da, da (vezi mesajul de mai sus, din 25 mai) :)
Din pacate greva asta ne cam da peste cap  :(

Thanks pentru informatie.Eu locuiesc la Roman ,si nu stiam daca sa plec vineri sau joi.pentru ca eu fac in jur de 5h pana la CAMPULUNG MOLDOVENESC.

Eu din pacate fac 12h in mod normal, adica cu tren direct  :(

Stiu ca e un topic mai vechi, dar m-am gandit sa contribui si eu putin, acesta fiind primul meu post pe acest forum... :)

Aceasta a fost o initiativa minunata, din pacate nu toate scolile au fost dispuse sa dea banii respectivi...Eu, de exemplu, am luat in 2005, in a VII-a, premiul I la ONM cu maxim de punctaj, dar nu m-am calificat in lotul pentru OBMJ si ar fi parut normal sa ma aflu si eu la acel "lot al sperantelor". Din pacate, scoala mea n-a vrut sa plateasca pentru mine...E trist, pentru ca as fi fost dispus sa dau din banii mei pentru a participa la acel lot dar, asta e, regulile au fost reguli...

Ptr. stormraven

Puteai sa le dai bani scolii si ea sa plateasca pentru tine! Apropo de la scoala din Cluj esti? Asa de curioziatate..

Da...de la Lic. Teoretic "Avram Iancu".
Presupun ca eram prea cascat ca sa ma gandesc la varianta asta. Oricum, eu ma gandeam ca regulie sunt stricte si nu se admit portite...


This isn't too difficult but is a fun thing to know.

The perpendicular bisectors of two of the side of a triangle ABC intersect the third side at the same point. Prove that ABC is a right triangle.

This is simple...Anyways - the answer.In $\Delta ABC$ , the circumcenter is the point $say , D $ on the hypotenuse.Now , since the circumcenter is equidistant from all the vertices, $AD = DC.$ Now , since the circumcenter lies on the midpoint of the hypotenuse , $\Delta ABC$ is a right $\Delta$


[quote="shobber"]Let $x,y,z$ be positive numbers such that
\[ {1\over x}+{1\over y}+{1\over z}=1.  \]
Show that
\[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}  \][/quote]

From the condition we have $\sum xy= xyz$
$S=\sum \sqrt{x+yz} \geq \sqrt{(\sum {\sqrt{x}})^2+(\sum \sqrt{yz})^2}$
Put $u=\sum \sqrt{x}$ $v=\sum \sqrt{yz}$
$\sqrt{u^2+v^2} \geq \sqrt{xyz}+u$
$u^2+v^2 \geq xyz +u^2+2\sum \sqrt{x}\sqrt{xyz}$
$\sum xy + 2\sqrt{xyz}\sum \sqrt{x} \geq xyz +2\sqrt{xyz} \sum \sqrt{x}$
But we have $\sum xy =xyz$  :lol:

Isn't that just walking forwards and going backwards?  :huh:

Aren't those equivalences, Karth? That would make the proof valid.

[quote="shobber"]Let $x,y,z$ be positive numbers such that \[{1\over x}+{1\over y}+{1\over z}=1.\] Show that \[\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\] [/quote]

Let $\sum = \sum_{\textrm{cyclic}}$. Square both sides to get the equivalent \[\sum (x+yz)+2\sum\sqrt{(x+yz)(y+zx)}\geq (xyz+x+y+z)+2\left(\sum (x+1)\sqrt{yz}\right)\] It's given that $xy+yz+zx=xyz$, so we just want to prove that \[\sum\sqrt{(x+yz)(y+zx)}\geq \sum (x+1)\sqrt{yz}\] But if we could prove that $\sqrt{(x+yz)(y+zx)}\geq (z+1)\sqrt{xy}$, we would be done. This inequality is equivalent with $xy+x^{2}z+y^{2}z+xyz^{2}\geq xyz^{2}+2xyz+xy$, i.e. $z(x-y)^{2}\geq 0$, which is obvious.

[quote="shobber"]Let $x,y,z$ be positive numbers such that \[{1\over x}+{1\over y}+{1\over z}=1.\] Show that \[\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\] [/quote]

$\sum \sqrt{x+yz}= \sum \sqrt{x \frac{xyz}{xy+yz+zx}+yz}$
$=\sum \sqrt{\frac{yz}{xy+yz+zx}(x^{2}+xy+yz+zx)}$
$=\sum \sqrt{\frac{yz(x+y)(x+z)}{xy+yz+zx}}$
$=\frac{1}{\sqrt{xy+yz+zx}}\sum \sqrt{yz(x+y)(x+z)}$
$\geq \frac{1}{\sqrt{xy+yz+zx}}\sum \sqrt{yz(x+\sqrt{yz})^{2}}$
$= \frac{1}{\sqrt{xy+yz+zx}}\sum (x\sqrt{yz}+yz)$
$=\sum (\sqrt{\frac{x^{2}yz}{xy+yz+zx}}+\frac{yz}{\sqrt{xy+yz+zx}})$
$=\sum \sqrt{x}+\frac{\sum{yz}}{\sqrt{xy+yz+zx}}$
$=\sum \sqrt{x}+\sqrt{xy+yz+zx}$
$=\sum \sqrt{x}+\sqrt{xyz}$.

The problem duces to $a+b+c=1$, and a problem solved by Minkowski ineq.

Let $a=\frac{1}{x}$, $b=\frac{1}{y}$, $c=\frac{1}{z}$. Then, $a+b+c=1$
$\sum = \sum_{cyc}$

$LHS = \sum \sqrt{x+yz}= \sum \frac{\sqrt{bc+a}}{\sqrt{abc}}= \sum \frac{bc+a(a+b+c)}{\sqrt{abc}}= \sum \frac{\sqrt{(a+b)(a+c)}}{\sqrt{abc}}\geq \sum \frac{a+\sqrt{bc}}{\sqrt{abc}}= \sum \sqrt{xyz}(\frac{1}{x}+\frac{1}{\sqrt{yz}}) = \sqrt{xyz}\sum \frac{1}{x}+\sum \sqrt{x}= \sqrt{xyz}+\sum \sqrt{x}= RHS$

Is it the same as your Minkowski ineq? If yes then I am sorry.

Does this work? PS sorry for bringing up an old topic.

[b]Solution[/b]

By Cauchy-Schwarz, it follows that for any $ x,y,z > 0$,

$ (x \plus{} yz)(y \plus{} xz) \ge (\sqrt {xy} \plus{} z \sqrt {xy})^2$
$ \sqrt {(x \plus{} yz)(y \plus{} xz)} \ge \sqrt {xy} \plus{} z \sqrt {xy}$

Summing this for all pairings of $ x,y,z$, it follows that

$ \sqrt {(x \plus{} yz)(y \plus{} xz)} \plus{} \sqrt {(x \plus{} yz)(z \plus{} xy)} \plus{} \sqrt {(z \plus{} xy)(y \plus{} xz)} \ge \sqrt {xy} \plus{} \sqrt {xz} \plus{} \sqrt {yz} \plus{} z \sqrt {xy} \plus{} x \sqrt {yz} \plus{} y \sqrt {xz}$

Now note that by the condition, $ xy \plus{} xz \plus{} yz \equal{} xyz$. Multiplying by $ 2$ and adding $ x \plus{} y \plus{} z \plus{} xy \plus{} xz \plus{} yz \equal{} x \plus{} y \plus{} z \plus{} xyz$ gives that,

$ (\sqrt {x \plus{} yz} \plus{} \sqrt {y \plus{} zx} \plus{} \sqrt {z \plus{} xy})^2 \ge (\sqrt {xyz} \plus{} \sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z})^2$

The result follows.

$ \sqrt{x\plus{}yz}\plus{}\sqrt{y\plus{}zx}\plus{}\sqrt{z\plus{}xy}\ge\sqrt{xyz}\plus{}\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}$
$ \iff \sum_{cyc} \sqrt{\frac{1}{yz}\plus{}\frac{1}{x}} \ge 1\plus{} \sum_{cyc} \frac{1}{\sqrt{yz}}$
$ \iff \sum_{cyc} \sqrt{(\frac{1}{x}\plus{}\frac{1}{y})(\frac{1}{x}\plus{}\frac{1}{z})} \ge 1\plus{} \sum_{cyc} \frac{1}{\sqrt{yz}}$ (since $ (\frac{1}{x}\plus{}\frac{1}{y})(\frac{1}{x}\plus{}\frac{1}{z})\equal{}\frac{1}{yz}\plus{}\frac{1}{x}$)
By C-S inequality,
$ \sum_{cyc} \sqrt{(\frac{1}{x}\plus{}\frac{1}{y})(\frac{1}{x}\plus{}\frac{1}{z})} \ge \sum_{cyc}(\frac{1}{x}\plus{}\frac{1}{\sqrt{yz}} )\equal{}1\plus{} \sum_{cyc} \frac{1}{\sqrt{yz}}$ ( since $ {1\over x}\plus{}{1\over y}\plus{}{1\over z}\equal{}1$)
QED

After dividing by $\sqrt{xyz}$ and making the substitution $x=\frac{1}{a}, y=\frac{1}{b}$ and $z=\frac{1}{c}$, the inequality becomes
\[ \sqrt{a+bc} +\sqrt{b+ca} +\sqrt{c+ab}\geqslant 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca} \]
Notice that $a+bc=(a+b)(a+c)$ and by Cauchy Schwarz $(a+b)(a+c)\geqslant (a+\sqrt{bc})^2$. Using this we have

\[ \sqrt{a+bc} +\sqrt{b+ca} +\sqrt{c+ab}\geqslant a+\sqrt{bc} +b+ \sqrt{ca} +c +\sqrt{ab}  \]
We're done due to $a+b+c=1$.

The condition gives
\begin{align*}
xyz = xy + xz + yz \iff yz + x = x + y + z + \frac {yz} {x}
\end{align*}
Furthermore, $\sqrt {xyz} = \left (\dfrac {1} {x} + \dfrac {1} {y} + \dfrac {1} {z}\right) \sqrt {xyz} = \sqrt {\dfrac {xy} {z}} + \sqrt {\dfrac {xz} {y}} + \sqrt {\dfrac {yz} {x}}$, so it suffices to show
\begin{align*}
\sum \sqrt {\dfrac {yz} {x} + x + y + z} \ge \sum \sqrt {\dfrac {xy} {z}} + \sum \sqrt {x}
\end{align*}
From AM-GM,
\[\dfrac {xy} {z} + x + y + z \ge \dfrac {xy} {z} + z + 2 \sqrt {xy} \implies \sqrt {\dfrac {xy} {z} + x + y + z} \ge \sqrt {\dfrac {xy} {z}} + \sqrt {z}\] Summing cyclically gives the desired result.

It's equivalent to show \[ \sum_{\text{cyc}} \sqrt{x+yz\left( \frac1x+\frac1y+\frac1z \right)} \ge \sum_{\text{cyc}} \sqrt{x} + \sqrt{xyz} \cdot \frac1x \] but now we're done by Cauchy as \[ \sqrt{x+yz\left( \frac1x+\frac1z+\frac1z \right)}= \frac{\sqrt{(x+y)(x+z)}}{\sqrt{x}} \ge \frac{x+\sqrt{yz}}{\sqrt{x}} = \sqrt x + \frac{\sqrt{xyz}}{x}. \]

The inequality is equivalent to
Let $a, b ,c$  be positive real numbers such that $a+b+c=1$ , prove that\[
\frac{a}{\sqrt{a+bc}+\sqrt{bc}}+\frac{b}{\sqrt{b+ca}+\sqrt{ca}}+\frac{c}{\sqrt{c+ab}+\sqrt{ab}}\ge 1.\]
[quote="vyfukas"]We have that $b+c=1-a \Rightarrow 1-2a+a^2=b^2+2bc+c^2\ge 4bc \Rightarrow bc \le \dfrac{1-2a+a^2}{4}$
Then we substitute this result in a given inequality:
$\dfrac{a}{\sqrt{a+bc}+\sqrt{bc}}\ge\dfrac{a}{\sqrt{a+\dfrac{1-2a+a^2}{4}}+\sqrt{\dfrac{1-2a+a^2}{4}}}=\dfrac{a}{\dfrac{a+1}{2}+\dfrac{1-a}{2}}=a$
Then adding similar inequalities we will get $a+b+c=1$ thus we are done.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=568982

[quote=kapilpavase]Since $xyz=xy+yz+zx$ we see that the individual terms on the RHS and LHS are equal when squared.So after squaring the ineq,it is sufficient to prove $$\sum \sqrt{(x+yz)(y+xz)}  \ge \sum \sqrt{xy} +\sum z\sqrt{xy}$$ which is Cauchy schwarz in 'most normal form' :P[/quote]

I am afraid, your solution is not different enough from post #8 to bring up such an old topic. ;)

[quote=shobber]Let $x,y,z$ be positive numbers such that ${1\over x}+{1\over y}+{1\over z}=1.$ Show that
\[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \][/quote]
http://www.artofproblemsolving.com/community/c6h568982p3338102
Let $x,y,z$ be positive numbers such that $x+ y+ z=1.$ Show that
\[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \]
[url=http://www.artofproblemsolving.com/community/c6h1201465p5909023]Stronger than APMO 2002[/url]

http://www.artofproblemsolving.com/community/c6h133953p758077
If $ x$, $ y$, $ z$ are three nonnegative reals, then prove that$$ \sum_{\text{cyc}}\sqrt {\left(z + x\right)\left(x + y\right)}\geq x + y + z + \sqrt3\cdot\sqrt {yz + zx + xy}$$
http://www.artofproblemsolving.com/community/c6h200514p1102800
Prove if $a,b,c>0$ and $ab+bc+ca=3$ then
$$ \sum\sqrt{a^2\plus{}3}\geq a\plus{}b\plus{}c\plus{}3$$

[hide = Quick]
Substitute $a = \frac{1}{x}, b = \frac{1}{y}, c = \frac{1}{z}$. Then $a + b + c = 1$. Then inequality we wish to prove becomes
$$\sum_{cyc}^{}\sqrt{\frac{1}{a}+\frac{1}{bc}} = \sum_{cyc}^{}\sqrt{\frac{a+bc}{abc}} \ge \sqrt{\frac{1}{abc}} + \sum_{cyc}^{}\sqrt{\frac{bc}{abc}}$$
Multiplying both sides by $\sqrt{abc}$, it suffices to prove
$$\sum_{cyc}^{}\sqrt{a+bc}\ge 1 + \sum_{cyc}^{}\sqrt{ab} = \sum_{cyc}^{}(a+\sqrt{bc})$$
If we can prove $\sqrt{a+bc} \ge a + \sqrt{bc}$, then we will be done. Squaring, we obtain
$$a+bc \ge a^2 + bc + 2a\sqrt{bc}$$
$$a \ge a^2 + 2a\sqrt{bc}$$
$$1 \ge a + 2\sqrt{bc}$$
$$1 - a = b + c \ge 2\sqrt{bc}$$
but last inequality true by AM-GM. $\blacksquare$
[/hide]

Homogenise the inequality to
$$ \sum_\text{cyc} \sqrt{\frac{(x+y)(x+z)}{x}} \ge \sum_\text{cyc} \sqrt{x} + \sqrt{\frac{yz}{x}}$$
We show that $\sqrt{\frac{(x+y)(x+z)}{x}} \ge \sqrt{x} + \sqrt{\frac{yz}{x}}$. Upon squaring and expanding, it reduces to $y+z \ge 2\sqrt{yz}$ which is obvious by AM-GM.

[quote=WizardMath]
We show that $\sqrt{\frac{(x+y)(x+z)}{x}} \ge \sqrt{x} + \sqrt{\frac{yz}{x}}$. [/quote]
This is true by Cauchy, no need to square both the sides :-D

Yeah very true. Wizard math nice. Solution. In inequality expansion and getting is great

Karamata inequality can kill this problem immediately.

Let   $   a,b,c>0   $   and   $   \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1   $ . [url=https://artofproblemsolving.com/community/c6h1904068p13025765]Prove that[/url] $$\sqrt{\frac{a-1}{b}}+\sqrt{\frac{b-1}{c}}+\sqrt{\frac{c-1}{a}} \ge \sqrt{6}$$$$\iff$$$$a\sqrt{1+\frac{c}{b}}+b\sqrt{1+\frac{a}{c}}+c\sqrt{1+\frac{b}{a}} \ge \sqrt{6(ab+bc+ca)}$$
[url=https://artofproblemsolving.com/community/c6h539040p3100246]h[/url]

a different solution:
$$\sum \sqrt{x+yz} \geq \sqrt{xyz} +\sum \sqrt{x} \Leftrightarrow \sum \frac{yz}{\sqrt{x+yz}+\sqrt x} =\sum \sqrt{x+yz}-\sqrt{x}\geq\sqrt{xyz}$$
which is correct because:
$$\sum \frac{yz}{\sqrt{x+yz}+\sqrt x}=\sum \frac{yz}{\sqrt{x+yz(\frac 1x +\frac 1y +\frac 1z)}+\sqrt x}=\sqrt{xyz}\sum \frac{yz}{\sqrt{(xz+yz)(xy+zy)}+ x\sqrt{yz}}\geq  \sqrt{xyz}\sum \frac{yz}{\frac{xz+yz +xy +yz}{2} +\frac{xy+xz}{2}}=\sqrt{xyz}$$

Let $x,y,z$ be positive numbers such that $\frac{1}{x}+\frac{2}{yz}=1.$  Show that
$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\geq\frac{3\sqrt{3}}{5}(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} )$$

Note that $\sqrt{xyz} = \sqrt{\frac{yz}{x}} + \sqrt{\frac{zx}{y}} + \sqrt{\frac{xy}{z}}$.
Now Note that we need to prove $\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy} \ge \sqrt{\frac{yz}{x}} + \sqrt{\frac{zx}{y}} + \sqrt{\frac{xy}{z}}+\sqrt{x}+\sqrt{y} + \sqrt{z}$ so we will prove $\sqrt{x+yz} \ge \sqrt{\frac{yz}{x}} + \sqrt{x}$ or $x + yz \ge \frac{yz}{x} + x + 2\sqrt{yz}$. Note that $\frac{yz+zx+xy}{xyz} = 1 \implies yz = xyz - zx- xy \implies \frac{yz}{x} = zy - z - y$ so we need to prove $y + z \ge 2\sqrt{yz}$ which is obviously true.

The condition is equivalent to $xyz=xy+yz+zx$. We'll use this to homogenize the left-hand side of the inequality:
\[\sqrt{xyz}\sum\limits_{sym} \sqrt{x+yz}=\sum\limits_{sym} \sqrt{x\cdot xyz+yz\cdot (xy+yz+zx)}=\sum\limits_{sym}\sqrt{yz}\cdot \sqrt{(x+y)(x+z)}\]
Now note that using the Cauchy-Schwarz inequality, we have that:
\[\sqrt{xyz}\sum\limits_{sym} \sqrt{x+yz}=\sum\limits_{sym}\sqrt{yz}\cdot \sqrt{(x+y)(x+z)}\geq \sum\limits_{sym}\sqrt{yz}\cdot (x+\sqrt{yz})=\sqrt{xyz}\sum\limits_{sym} \sqrt{x}+xyz\sum\limits_{sym} \frac{1}{x}=\sqrt{xyz}\sum\limits_{sym} \sqrt{x}+xyz\]
which is equivalent to the original inequality after diving both sides by $\sqrt{xyz}>0$.

[quote=shobber]Let $x,y,z$ be positive numbers such that $ {1\over x}+{1\over y}+{1\over z}=1. $ Show that
\[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \][/quote]

[quote=shobber]Let $x,y,z$ be positive numbers such that
\[ {1\over x}+{1\over y}+{1\over z}=1. \]
Show that
\[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \][/quote]
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$${1\over x}+{1\over y}+{1\over z}=1$$
$$\Rightarrow xy+yz+zx=xyz...(i)$$
$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$
$$\Leftrightarrow x+yz+y+zx+z+xy+2\sum_{cyc} \sqrt{xy+x^2z+y^2z+xyz^2}\ge xyz+x+y+z+2\sum_{cyc}\sqrt{x^2yz}+\sqrt{xy}$$
By $(i):$
$$\Leftrightarrow \sum_{cyc}\sqrt{(x+yz)(y+xz)}\ge \sum_{cyc}\sqrt{xyz^2}+\sqrt{xy}$$
It is enough to prove that:
$$\sqrt{(x+yz)(y+xz)}\ge \sqrt{xyz^2}+\sqrt{xy}$$
$$\Leftrightarrow xy+x^2z+y^2z+xyz^2\ge xyz^2+xy+2\sqrt{x^2y^2z^2}$$
$$\Leftrightarrow (x^2+y^2)\ge 2xy$$
which is trivial by $AM-GM$
$$\Rightarrow \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}_\blacksquare$$
$\color{blue}\rule{24cm}{0.3pt}$


If a eukaryotic cell has a diameter that is 10 times that of a bacterial cell, proportionally how much more surface area would the eukaryotic cell have?

Proportionally how much more volume would it have?

I have the Campbell AP Biology textbook, but I don't get what it's doing when it uses the cube concept.

Thanks in advance!

Assuming they're both perfect spheres,

R=diameter of eukaryotic cell
r=diameter of bacterial cell

10r=R

The surface area of a sphere is $ 4\pi r^2$.

So ratio of eukaryote cell's surface area to bacterial cells surface area would just be a ratio of their radii squared (the $ 4\pi$ would cancel out), and since 10r=R, R/r=10, therefore $ R^2/r^2 \equal{} 100$.

The volume of a sphere is $ 4\pi r^3/3$.
Similarly for volume, it'd be the ratio of their radii cubed (the $ 4\pi/3$ cancels out), so R/r=10, therefore $ R^3/r^3 \equal{} 1000$.

EDIT: Sorry, didn't see this post was made in August '09. >.<


[quote="Sir.Aaron"]Find the area of the  figure. Dimensions are in feet

[img]http://i21.photobucket.com/albums/b276/Toxic590/untitled.jpg[/img][/quote]

[hide]91[/hide]

[hide]Area of rect.=7x9=63
Area of tri.=7x8/2=28
Area of figure=63+28=91[/hide]

how about the area of this figure

[img]http://i21.photobucket.com/albums/b276/Toxic590/untitled-1.jpg[/img]

[hide]the total area = area of trapezoid - area of semicircle
area = $264-32\pi$[/hide]

The First figure-91
The Second one-264-32pi

Split the figure into 3 rectangles whose dimensions are:
$2$ by $3$, $2$ by $3$, and $3$ by $3$
It's easy to find the area now:
$2\cdot 3+2\cdot 3+3\cdot 3 = 21$

Thanks for all your help guys. I really appreciate it. Iv got one more problem, how do I find the shaded portion of this figure?

[img]http://i21.photobucket.com/albums/b276/Toxic590/kkk.jpg[/img]

I assume that the circle is tangent to the bases of the trapezoid. The median of the trapeziod is $\frac{6+8}{2}=7$. The height of the trapeziod is the circle's diameter, of $4$. So the area of the trapeziod is $7\cdot 4 = 28$. The area of the circle is $\pi 2^{2}= 4\pi$. The shaded region's area is the trapeziod's area minus the circle's area, or $28-4\pi$.

[hide]Area of trapezoid 4*14/2=28

Area of circle 2^2*pi=4i

so answer is 28-4pi.[/hide]

[quote="Sir.Aaron"]How about the area of this one? All angles are right angles

[img]http://i21.photobucket.com/albums/b276/Toxic590/untitledL.jpg[/img][/quote]

30-9=21


$ \prod_{k\equal{}1}^\infty 1\minus{}\frac{1}{3^k}$
maybe it's possible to generalize?
$ \prod_{k\equal{}1}^\infty \frac{a^k\minus{}b}{a^k}$

[quote="Mgccl"]$ \prod_{k \equal{} 1}^\infty 1 \minus{} \frac {1}{3^k}$
maybe it's possible to generalize?
$ \prod_{k \equal{} 1}^\infty \frac {a^k \minus{} b}{a^k}$[/quote]

Uhh, you're multiplying an infinite number of fractions... isn't it theoretically 0?

Infinite products and series can converge to things other than $ \pm \infty$ and $ 0$ under certain conditions. In this case notice that $ 1 \minus{} \frac{1}{3^k}$ tends to $ 1$ (the multiplicative identity) for large values of $ k$.

The bad news is, there is no "nice" form for that answer (correct me if I'm wrong).

However, we can show that it converges to a positive number, which I will leave as an open exercise for anyone interested

Let $ a_n \equal{} 1 \minus{} \frac{1}{3^n} \equal{} \frac{3^n \minus{} 1}{3^n}$, and $ a_n > 0$ for $ n \geq 1$. Now for $ P \equal{} \prod b_n$ then $ \ln P \equal{} \sum \ln(b_n)$, and so P converges iff ln P converges. For $ b_n \equal{} a_n$, the sum $ \sum_{n\equal{}1}^{\infty}\ln\left(1 \minus{} \frac{1}{3^n}\right)$ converges, because

$ \lim_{n \to \infty} \frac{\ln\left(1 \minus{} \frac{1}{3^n}\right)}{\frac{1}{3^n}} \equal{} \minus{}1$

and $ \sum_{n\equal{}1}^{\infty} \frac{1}{3^n}$ converges. Therefore, $ \prod_{n\equal{}1}^{\infty} \left(1 \minus{} \frac{1}{3^n}\right)$ converges also. Its value is approximately 0.560126 (look entry number 48 [url=http://mathworld.wolfram.com/InfiniteProduct.html]here[/url]).


limit
calculus
derivative
function
inequalities
real analysis
real analysis unsolved

Here is the problem from the book Sadovnichiy, Podkolzin"USSR student's olympiads" that i think is uncorrect:
   Let $x_{k+1}=\frac{1}{2}(x_k+\frac{a}{2x_k^2})$ where $a>0$  and $x_0>0$.
(i) Prove that $\lim x_n$ exist
(ii) let $y_n$ be the differnce between $x_n$  and $\lim x_n$ and $y_0$ is not equal to zero. Prove:
     $y_n>0$
$y_{n+1}<\frac{2}{3}y_n$
$y_{n+1}<\frac{1}{\sqrt[3]a}y_n$

I think that it must be $x_{k+1}=\frac{1}{3}(x_k+\frac{a}{2x_k^2})$ but i cant figure out the $y_{n+1}<\frac{1}{\sqrt[3]a}y_n$

 :huh:

Hey help be anybody with $y_{n+1}<\frac{1}{\sqrt[3]a}y_n$?!

Well, the problem is that even the first assertion $y_n>0$ is false unless the coefficient is $\frac 2 3$ (not $\frac 12$ or $\frac 13$). The coefficient $\frac 13$ is really bad because then the limit, if it exists at all, must be $\sqrt[3]{\frac a4}$ but the derivative of the function $\frac13\left(x+\frac a{2x^2}\right)$ at this point is $-1$, which means that even if the convergence takes place, it is extremely slow.  Note also that if $x_0$ is close to $0$, than $y_0$ is, roughly speaking, a fixed number but $y_1$ is as large as we want. So, it might be a good idea to figure out the right formulation of the statement before trying to prove anything here.

I'm sorry that wrote by a mistake $\frac{1}{3}$. I solved the problem when $\frac{2}{3}$ AND CANT FIGURE OUT THE LAST INEQUALITY. please help me, it is taking a lot of my expencive time. :blush:

It is still false without extra assumptions: take $a$ very large and  $x_1$ really huge. Then $y_1\approx x_1$. Now, $x_2\ge \frac 2 3x_1$ is also huge, so $y_2\approx x_2 \ge \frac 2 3 x_1>\frac 2 3y_1$. So the improvement in the error is only by the factor of $\frac 2 3$, not by that of $\frac{1}{\sqrt[3]a}$. And this remains true as long as $x_k$ stay much greater than the limit $\sqrt[3]a$.

Seems like I figured out what the last part should really be: one needs to prove that $y_{n+1}\le \frac 1{\sqrt[3]a}y_n^2$  :lol:

Thanx a lot fedja. Now I know how should be the problem really and the proof of the last inequality is easy :)


inequalities
algebra
polynomial
inequalities unsolved

I don't understand the Chinese in [url]http://www.chinamath.cn/manage/wenzhang/sxxb-2006-49(3)-491.pdf[/url]. Do you mean that any symmetric polynomial in $a,b,c$ is semidefinite iff it can be decomposed into sums of "schur sums" of the four types
 $\sum x^{k}(x-y)(x-z) \ge0 (k \ge 0),$  (2.2.11)
 $\sum (yz)^{k}(x-y)(x-z) \ge0 (k \ge 0),$ (2.2.12)
 $\sum x^{k}(y+z)(x-y)(x-z) \ge0 (k \ge 1),$ (2.2.13)
 $\sum (yz)^{k}(y+z)(x-y)(x-z) \ge0 (k \ge 0).$ (2.2.14)

with non-negative coefficients?


An odd numbre is a number that can be written on the form 2k+1 for any k that belongs to the set of natural numbers (including 0)

[quote="Cauchy_Schwartz"]Can someone tell me what is an odd natural number? Please! thank you very much![/quote]

I think this is a question about the English language. The odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 . . .

[quote="Cauchy_Schwartz"]Can someone tell me what is an odd natural number?[/quote]

I think this is a question about the English language. The odd natural numbers are the numbers in the set {1, 3, 5, 7, 9, 11, 13, 15, 17 . . . }.

Thank you very much! And now could you tell me how do you call 2, 4, 8,......, 2k?

Ok, thank yo very much! You dont have to be so hard on me...

Odd = not multiple of 2
Even = multiple of 2

Sorry Cauchy_Schwartz, didn't mean to be hard on you. Just wanted to emphasize the word even.

One of my colleagues from France knew the English words "odd" and "even", but he had trouble remembering which was which.  I tried to help him come up with a mnemonic, and then we realized that "odd" has an odd number of letters, while "even" has an even number of letters.  After that observation, he never forgot.

That's a great ideea. Thanks Ravi B...I only knew that odd=weird

Legend has it that Famous  Enrico Fermi when he just came to USA used to   talk about "even" and "uneven" functions in his Physics lectures.   He had some graduate students who  used to monitor his English language and   it is said that he was such    a good learner of   the anguage   that he would never make a language mistake  second time ( after it was pointed out by the graduate students)  but it seemed that the students, for a long time, never corrected him, becuase they thought  it was real fun...(and besides  everyone knew what he was talking about...)  ... and 'uneven' term was  fondly used by many of his colleagues.


Is there a polynomial $ P(x): \mathbb{R}\to \mathbb{R}^ \plus{} \cup \{0\}$ which satisfies the following three conditions.

1. $ P(n)\ne P(m)$ for every $ n,m\in\mathbb{Z},n\ne m$.
2. $ P(n)\in\mathbb{Z}^ \plus{} \cup \{0\}$ for every $ n\in\mathbb{Z}$.
3. For every $ m\in\mathbb{Z}^ \plus{} \cup \{0\}$ there exists $ n\in\mathbb{Z}$ such that $ P(n) \equal{} m$?

No.  The only polynomials that satisfy condition 3 are the polynomials of the form $ P(x) \equal{} x \plus{} a$.  (To see this, note that the finite differences of a polynomial of degree $ 2$ or higher are unbounded.)


Let x1,...xn be real positive numbers whose sum is not more than1/2.
Prove that

        (1-x1)(1-x2)...(1-xn) \geq 1/2

we may prove
(1-x1)(1-x2)...(1-xn) >= 1 - (x1+x2+...xn)
by mathematical induction.


function
inequalities
inequalities proposed

I strongly believe that yours is true, and mixing variables method will solve this. I have not tried it out yet.

[quote="pvthuan"]Prove that if $a,b,c>0$ then
\[ (a+b+c)\Bigl(\frac a{(a+b)^2}+\frac b{(b+c)^2}+\frac c{(c+a)^2}\Bigl)\geq\frac94.  \]
It is true for $a\geq b\geq c$. What about other cases?[/quote]

Well, in other cases it is not necessarily true; e. g., it is wrong for a = 0, b = 1, c = 2.

For $a\geq b\geq c$, it is rather easy to show:

[color=blue][b]Lemma 1.[/b] Let $f : I \to \mathbb{R}$ be a convex function on an interval $I \subseteq \mathbb{R}$, and let $a, b, c$ be three numbers from this interval $I$ satisfying $a\geq b\geq c$. Then,

$\left(b-c\right)f\left(a\right)+\left(c-a\right)f\left(b\right)+\left(a-b\right)f\left(c\right)\geq 0$.[/color]

[i]Proof of Lemma 1.[/i] We WLOG assume that $a > c$ (since otherwise, we have $a = b = c$, and thus the inequality in question boils down to $0 \geq 0$). Since the function $f$ is convex on the interval $I$, we have

$\frac{\left(b-c\right)f\left(a\right)+\left(a-b\right)f\left(c\right)}{\left(b-c\right)+\left(a-b\right)}\geq f\left(\frac{\left(b-c\right)a+\left(a-b\right)c}{\left(b-c\right)+\left(a-b\right)}\right)$

(note that the coefficients $b - c$ and $a - b$ are both $\geq 0$, since $a\geq b\geq c$). In other words,

$\frac{\left(b-c\right)f\left(a\right)+\left(a-b\right)f\left(c\right)}{a-c}\geq f\left(\frac{\left(a-c\right)b}{a-c}\right)=f\left(b\right)$.

Multiplying this by the nonnegative number $a - c$ (nonnegative since $a\geq c$), we get

$\left(b-c\right)f\left(a\right)+\left(a-b\right)f\left(c\right)\geq\left(a-c\right)f\left(b\right)=-\left(c-a\right)f\left(b\right)$,

and thus

$\left(b-c\right)f\left(a\right)+\left(c-a\right)f\left(b\right)+\left(a-b\right)f\left(c\right)\geq 0$.

Hence, Lemma 1 is proven.

Now, set $S = a + b + c$, and let $I$ be the interval $\left(0; S\right) \subseteq \mathbb{R}$. Consider the convex function $f : I \to \mathbb{R}$ defined by $f\left(x\right)=\frac{1}{\left(S-x\right)^2}$ on this interval, and the numbers $a, b, c$ from this interval. These numbers satisfy $a\geq b\geq c$. Hence, Lemma 1 yields $\sum\left(b-c\right)\frac{1}{\left(S-a\right)^2}\geq 0$, where the $\sum$ sign stands for cyclic sums. Since $S = a + b + c$, we have $S - a = b + c$, and thus this becomes $\sum\left(b-c\right)\frac{1}{\left(b+c\right)^2}\geq 0$, i. e. $\sum\frac{b-c}{\left(b+c\right)^2}\geq 0$. Thus,

$\sum\frac{b}{\left(b+c\right)^2}=\sum\frac12\cdot\frac{\left(b+c\right)+\left(b-c\right)}{\left(b+c\right)^2}$
$=\sum\frac12\cdot\left(\frac{1}{b+c}+\frac{b-c}{\left(b+c\right)^2}\right)=\frac12\cdot\sum\frac{1}{b+c}+\frac12\cdot\underbrace{\sum\frac{b-c}{\left(b+c\right)^2}}_{\geq 0}$
$\geq\frac12\cdot\sum\frac{1}{b+c}$.

But by Cauchy-Schwarz, $\sum\frac{1}{b+c}\cdot\sum\left(b+c\right)\geq\left(\sum\sqrt{\frac{1}{b+c}\cdot\left(b+c\right)}\right)^2=\left(\sum 1\right)^2=3^2=9$, so that $\sum\frac{1}{b+c}\geq\frac{9}{\sum\left(b+c\right)}=\frac{9}{2\left(a+b+c\right)}$, and thus

$\sum\frac{b}{\left(b+c\right)^2}\geq\frac12\cdot\sum\frac{1}{b+c}\geq\frac12\cdot\frac{9}{2\left(a+b+c\right)}=\frac{9}{4\left(a+b+c\right)}$,

so that $\left(a+b+c\right)\cdot\sum\frac{b}{\left(b+c\right)^2}\geq\frac94$, and your inequality is proven.

[quote="nthd"]I have a similar ineq ,but I can't solve it:$a,b,c>0$
Prove that
         $(a^2+b^2+c^2)(\frac{a^2}{(a^2+bc)^2}+\frac{b^2}{(b^2+ca)^2}+\frac{c^2}{(c^2+ab)^2} )\ge \frac{9}{4}$[/quote]

Yes, you posted that at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=43877[/url] , and it really looks very hard. If I will get some progress on this inequality, I will post it.

  Darij


Just wondering which round is more liked, sprint or target.

let's see...

sprint
_____
messup=ok
get one=ok

target
_____
messup=horrible
get one=great

so i decided the sprint is more liked

but target is risk

jorian

[quote="Moldytape"]I'm better at easyish problems with an average of 1 min 20 seconds per problem than easy or hard problems with 3 minutes per problem that if you mess up then your self esteem will drop. Also, there's more tension between the sprint round since you basically have absolutely no idea what you got unless you compare a lot of your answers to someone else who wrote everything in a scrap piece of paper. You can just talk to the other kids your sitting with and agree what you got for your targets.[/quote]
In my state they call out the answers right after, for each round.

When I was in 7th grade, I hated target round, but I now love target round. :maybe: I do better on Target Rounds maybe because I am awakened from my sleep by then.

I ALWAYS mess up target in real.  Last year state - 6, national - 3.  'nuff said.

Sprint = pretty ok.  Last year chap - 30, state - 27, nats - 22. then This year chap - 29 (lol, I got #5 wrong...how many cubic feet are in a cubic yard..)

I like target, if you make a mistake your more likely to catch it, so I think it's a more accurate indicator, of course with only 8 questions it's not the best.

I really don't like target.

My calculator isn't all fancy like yours so I end up doing all this hard manual logic work (and not inputting some numbers into a program).

Also, for some reason, I always freak out towards the end of most National/and some Sate rounds, and don't bother to re-read questions and make stupid mistakes.

On the other hand, I consistantly get 30s in Chapter Sprints, 28s in State Sprints, and 23-25 in National Sprints.

Target, not so much.

[quote="footballrocks41237"]I like targets I guess... I always bomb the sprint because time is more of an issue there and I make stupid mistakes...like $6\times4=32$ because I can't read my own handwriting..[/quote]

LOL

Ironically, I had the exact same mistake (for a school test).

I've got one heck of a grudge against target... :mad: It totally screwed me up this year on MC... I just like sprint better. Its easier (in a way)

[quote="mathcrazed"]I usually do good in sprints during practice, but suck during competitions.  And own at targets in competitions... so -_- dunno[/quote]

Same here. When I take sprints for practice I pwn them with 28s, 29s and 30s. But when I take them in real life I fail. Anyway targets are usually okay but they aren't very fun. Sprint has a lot more excitement involved.

Target round is the better one. Personally, I'm good at ciphering, and target is similar (kind of). I usually do well, so it is good.

I like target, but I always mess up at the actual competition!  Like when I was preparing for school (it's kind of hard to make the team at my school), I got 8s on target on like every practice but got 6 on the actual thing :mad:  Because of that I was on the B team, but then won my chapter  :rotfl:

i like target more, but i screwed them up so much this year. I think a mess-up on the target costed me a spot in the top ten at states... not to mention last year!

i got an 8 ot of sixteen on last year's target.

i got an 8 out of sixteen on the target two years ago.

i sucked in seventh grade. i had a better score in states in 6th grade than i did in seventh grade (and my rank in sixth grade was less than half that of my seventh grade rank!)

conclusion: i sucked in seventh grade. laugh it up.

i got all of them target questions correct, but I broke my own record... (yeah you know what i mean)

man....I was going to keep the same score....

I do horrible at Sprint always get lower than 25 and higher than 20, once i got 29 on this years' school, but after this and before this, I never got higher than 25..(I'm not that good of a mathematician) :blush:


geometry
trigonometry
circumcircle
trig identities
Law of Sines
Law of Cosines

Let be ABC a triangle whit A=45, B=75. On (BC consider D, and BC=2DC.
Find <BAD.

The calculations are tedious for me, so I'll just explain the procedure and someone else and figure it out.

[hide]Let $AB=1$. From the law of sines, ${BC=\frac{\sqrt{6}}{3}}$. Since $D$ is the midpoint of $BC$, ${BD=\frac{\sqrt{6}}{6}}$. Then use the law of cosines to find $AD$. Then use the law of sines to find $AD=\frac{\sqrt{6}}{6\sin{BAD}}$. Then solve for the measure of angle $BAD$. [/hide]

[quote="vishalarul"]The calculations are tedious for me. [/quote]
[color=green]Not so tedious if you assume C is between B and D[/color][hide]
Then if $BE \perp AC$ it is easy to show that E is the center of the circumcircle of $\Delta ABD$ and $\angle BAD=60^{o}$[/hide]
[color=red]For mychrom[/color][hide]Please read the replies to your messages. Another problem from youhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=122908 has a question from Farenhejt and not an answer :) [/hide]


Spring Baseball. Awesome

June Baseball. Yay

July Baseball. Ok....

August Baseball. Alright, when do the playoffs start?!?!?

September Baseball. Cant wait for football

October Baseball. Yay World Series...........football's on.

Red Sox win. yifman is HAPPY HAPPY HAPPY HAPPY HAPPY HAPPY HAPPY HAPPY.

[quote="yif man12"]Spring. Baseball. Awesome.

Didn't start off too well for my Sox, but they scored their first win today! And what's with Mariano? 4 straight blown saves against Boston! And getting booed in Yankee Stadium too! Wow...[/quote]

sort of like BK against the Yankees in the 2001 World Series where he blew 2 save opportunities and put the D-Backs on the verge of elimination.

GIANTS all the way!!! :P Hopefully, we can still be afloat without Barry Bonds. 1-1 with 160 games left. Hey, guess what? Giants have a closer now!!!

Whoa, those Cleveland Indians today - scored three runs to tie it up 5-5 in the top of the ninth, and then scored 6 more in the top of the 11th to secure their win. Woohoo!  :D And after Close Bob Wickman gave up a three run lead in the bottom of the ninth yesterday... :(


[hide]
Difference of two squares, $4xy \ge 25$, y is smallest when x = y.
$y^2 = \frac{25}{4}$
$y = \frac{5}{2}$
[/hide]

[hide]We simplify the left hand side to get 4xy>=25, so xy>=25/4.

Now, since x=<y, then let us assume x=y.

So 4y^2>=25

So y>=5/2.

So y=5/2.

$\frac{5}{2}$[/hide]

[hide]Take the square root of each side. You get $(x+y)-(x-y)\geq 5$ Simplifying yields $2y\geq 5$ and thus $\frac{5}{2}$ is our answer.[/hide]

[quote="frt"][quote="ragnarok23"]Take the square root of each side. You get $(x+y)-(x-y)\geq 5$[/quote]
$\sqrt{(x+y)^2-(x-y)^2}\ne (x+y)-(x-y)$ ;)[/quote]
Shoot, I knew that. :D

LaTeX is taking forever to load on my computer I will try again. In the meantime, could a mod please change this into symbols and whatnot? Thanks.
[hide="My Beautiful Answer"] I took the square root of each side and got root (x+y)^2 - (x-y)^2 > root 25. Simplifying it is (x+y) - (x-y)>5. The x's cancel out so you get 2y>5. Divide by 2 to get y>5/2 (Please change the greater than signs into greater than or equal to whoever changes this, thanks). The answer is 5/2.[/hide]

Did this problem come from a practice SAT or a real SAT?

[hide]\begin{eqnarray*} (x+y)^2-(x-y)^2 &\geq& 25 \\ (x^2+y^2+2xy)-(x^2+y^2-2xy) &\geq& 25 \\ 4xy &\geq& 25 \\ xy &\geq& \frac{25}{4} \end{eqnarray*}
Because $0 \leq x \leq y$, I think the least value of y occurs when $x=y$ and $xy=25/4$, so $x=y=\sqrt{\frac{25}{4}}=\frac{5}{2}$[/hide]

Many of the posts have assumed that the minimum $y$ will occur when $x = y$ without sufficient rigor.  I would instead continue as follows.

[hide]As others have shown, the main inequality is equivalent to $xy \ge \frac{25}{4}$.  Because $y \ge x$, we have
\[ y^2 \ge xy \ge \frac{25}{4} \, .  \]
Taking square roots gives $y \ge \frac{5}{2}$.

The example $x = y = \frac{5}{2}$ shows that $y = \frac{5}{2}$ can be achieved.
[/hide]

[quote="Silverfalcon"]Normally, I [i]hate[/i] SAT math problems because they do not give fun you get from competitional problems. But this one was pretty fun to do. It's not very difficult but it was the last problem in the math section (grid-in problem).

If $0 \leq x \leq y$ and $(x + y)^2 - (x - y)^2 \geq 25$, what is the [u]least[/u] possible value of $y$?[/quote]

wow...an SAT problem that is hard enough for getting started  :D

SAT on AoPS... looks like the CollegeBoard is starting to learn...

[hide]
\[ (x+y)^2-(x-y)^2\geq 25 \\(x+y+x-y)(x+y-x+y)\geq 25 \\(2x)(2y)\geq 25 \\4xy=25 \\ \ Smallest \ y = x \\4y^2= 25 \\2y=5 \\y=\frac{5}{2} \\y=2.5 \]
[/hide]

[quote="Ravi B"]For the folks who are assuming $y = x$, here is a variant problem.

Problem 2: If $0 \le x \le y$ and $3y \ge x^2 + 2$, what is the least possible value of $y$?[/quote]
[hide]as x increases, y increases, so we want to use the smallest value of x, which is 0.  so $y=\frac{2}{3}$[/hide]

For those who care:

I remember this problem from the SAT preparation packet. My opinions on Collegeboard's math changed in seconds  :D

Too bad the same difficulty of problems wasn't on the real SAT.

[quote="236factorial"][quote="perfectnumber628"]random side note: when i took the sat a few months ago, there was this one question that required you to know $\pi=3.something$.  and it was "hard" compared to the rest of the math questions.[/quote]

pi key on your calculator...[/quote]
true.  ah, and i thought there was an sat math question that you actually have to know something for  ;)


The following figures are composed of squares and circles.
Which figure (A, B, or C) has a shaded region with largest area?
[asy]/* AMC8 2003 #22 Problem */
size(3inch, 2inch);
unitsize(1cm);
pen outline = black+linewidth(1);
filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);
filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);
filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));
filldraw(Circle((1,1), 1), white, outline);
filldraw(Circle((3.5,.5), .5), white, outline);
filldraw(Circle((4.5,.5), .5), white, outline);
filldraw(Circle((3.5,1.5), .5), white, outline);
filldraw(Circle((4.5,1.5), .5), white, outline);
filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);
label("A", (1, 2), N);
label("B", (4, 2), N);
label("C", (7, 2), N);
draw((0,-.5)--(.5,-.5), BeginArrow);
draw((1.5, -.5)--(2, -.5), EndArrow);
label("2 cm", (1, -.5));

draw((3,-.5)--(3.5,-.5), BeginArrow);
draw((4.5, -.5)--(5, -.5), EndArrow);
label("2 cm", (4, -.5));

draw((6,-.5)--(6.5,-.5), BeginArrow);
draw((7.5, -.5)--(8, -.5), EndArrow);
label("2 cm", (7, -.5));

draw((6,1)--(6,-.5), linetype("4 4"));
draw((8,1)--(8,-.5), linetype("4 4"));[/asy]

In Region A, the circle inside has radius 1, so the area of the circle is $ 1^2 * \pi \equal{} \pi$
Subtracting this from the the whole region of A, the shaded part is:

$ 4 \minus{} \pi$

Similarly, each of the circles in Region B has radius 1/2, so the area of each of the circles is $ (\frac {1}{2})^2*\pi \equal{} \pi/4$, and the combined areas of the circles is $ \pi$. So the area of the shaded part is:

$ 4 \minus{} \pi$

For Region C, the outer circle has area $ 1^2 * \pi \equal{} \pi$. The square inscribed in the circle has radius 1, and because it can be split into four 45-45-90 triangles, a side of the square is $ \sqrt {2} * 1 \equal{} \sqrt {2}$. So, the shaded area is $ \pi \minus{} \sqrt {2}^2 \equal{}$

$ \pi \minus{} 2$.

$ 4 \minus{} \pi$ is around .86, and $ \pi \minus{} 2$ is around 1.14

Thus, Region $ \boxed{C}$ has the shaded region with the largest area.

[quote="pi man"]Do you actually have to compute each figure's shade area? Isn't there another way?[/quote]
Other than just eyeing it (which I don't recommend you do), I'm pretty sure there is no other way to determine which shaded area is the greatest.  :wink:


AMC USA(J)MO USAMO ratio algebra polynomial function \/closed

I was considering buying this book, but I am not sure.

1. It was written by Douglas Faires so does this cover only AMC 10 or also AMC 12? (particularly the harder ones on 12) Does it show how to crack many AMC type problems like approaches and stuff or is it just general problem solving type of book?

2. I have read AOPS 1 and most of 2, would this book still be useful for raising AMC 12 scores?

3. Does this book have solutions to problems?

4. Is there an outline of the topics it covers... it just says 18 chapters.

5. Right now, I get 110-120 on AMC 12 (I haven't done any practice tests yet), would this book help me get 130+, or would practice tests be enough.

6. Two other books I was considering was mathematics by choice by ivan niven and 500 mathematical challenges. My goal is to qualify for USAMO, any suggestions of which of the 3 books?

Thanks!

Here I think I can answer some of your questions

1. This book covers mostly AMC 12 material rather than AMC 10 stuff. They mostly have problems from the AHSME. It also gives you solutions in the same way you get them for a AMC exam.

2. I think this book could help raise your AMC 12 score.

3. As I said earlier it does.

4. The topics it covers are mostly arithmetic ratios, polynomials, exponents, functions, geometry, polygons, counting, probability, prime decomposition, number theory, sequencies/series, stat, trig, 3d geometry, logs, and complex numbers. (I know that's not 18 because i mixed things like circle and triangle geometry together.)

5. I think you should go over this book plus some practice tests.

6. can't help you with those books since I've never read them.

Hope that'll help.


geometry quadratics parameterization algebra

so who is pumped...

i am sure all juniors are ready to bust out f2 solve...

Oh dude I should probably try to find you at some point oh well my extremely solid plan to find people is to wander about aimlessly around the quad and union.

Grr

I wanted to be on F/S 8-person team. Our current one is...ummm not too great and has a good chance of getting destroyed.   :maybe:

hmm i will be wearing a bright orange shirt along with the other people from my school who are man enough to wear such a shirt

we need like an aops pow wow

Naperville North will win.  Although I kinda hope we don't.  But I [i]guess[/i] I won't sabotage us.

man written owned me...i pretty much careless errored myself out of a metal...like i had a 34, but i lost on tiebreaks...

Oh man check out this careless mistake:

The problem said to round to 4 significant figures, and instead I rounded to 4 decimal places (which made my answer have 6 significant figures).  :mad:

Ummm ok so for some reason I decided to do 80% of my calculations on the individual thing by hand... also man we almost won the 2-person thingy... if they had given a "hard" geometry problem as the last question, then we would have won ;)

and uhh... good job rnwang except not really why aren't you doing the kinds of problems you're supposed to be doing... get away from that ICTM junk

also dude I find it pretty funny how intense some of these schools are about this silly ICTM stuff... like the guys who got last place on 2-person 3AA demonstration thing were pretty intense... maybe this is why IL is failing at MOP representation

Yes well now that ICTM is over, I can start doing intense preparation for AIME/USAMO and hopefully make it to MOP next year (which is actually pretty unlikely, but I will do the intense prep anyway just for fun  :roll: ).

well like ictm/nsml is like learn definition/plug numbers in

one nice problem on the 8 person [stolen from somewhere else of course...i think i saw this in the russian oly book]...too bad i was like wtf blank during the contest

solve $x=\sqrt{k-\sqrt{k+x}}$ provided $k>1$

[hide]ono solve for k[/hide]

[quote="Geometry Written 2007"]Using a compass only to construct circles or arcs of circles and an unmarked straightedge only to draw lines or rays or segments, give the letter(s) corresponding to all of the following that do [b]not[/b] determine a unique figure that can be constructed under the conditions previously noted. For your answer, write all the letters that apply. For example, if in a given set, all except G and K are uniquely determined and can be constructed, you would write: [b]G, K[/b] as your answer.

A. Construction of the positive mean between two given segments
B. Construction of a circle given 3 points on the circle
C. Construction of a segment whose length is the fourth proportional to the lengths of three given segments
D. Construction of a triangle given the lengths of the three medians of the triangle
E. Trisecting a given angle of unknown measure
F. Constructing an angle with a degree measure of 30
G. Constructing a segment of length $\frac{\sqrt{2}}{2}$ given a segment of length 1
H. Constructing a regular nonagon inscribed in a given circle[/quote]k wtf is a fourth proportional? :huh: mathworld and wikipedia do not know. i heard that like 1~2 students in the entire contest got this question right.. the [hide="answer was"]E,H[/hide] how would you do D?

Wow yeah I got that question wrong...I put D and E. :maybe:

what ...E, H obviously...C is probably just like a/b=c/x, or something...obviously constructable

Oh yeah, that $x=\sqrt{k-\sqrt{k+x}}$ for $k > 1$ problem was pretty cool. I ended up spending like 10 minutes doing it (basically square, isolate radical, square again, get quartic, factor quartic, solve quadratics, eliminate extraneous roots), but we won 8-person anyways :)

i am pretty sure i have seen that problem either in the russian math oly book, math olympiad treasures, or one of the random algebra problem books that the MAA publishes...when i get home i might look for it

but this is one of those: "solve for the parameter" problems...like y=5k...in some sense, the problem is contrived...but you know that it factors into quadratics...

haha there we go

i also hear that they allow graduating students to propose problems...i want to do that...

like ictm is okay i guess...but the "hard" problems are generally lame...like that one problem which was like find the integer roots of that large polynomial...

like at least help with the geometry test

[quote="ThAzN1"]but we won 8-person anyways :)[/quote]

Screw you Tony =)

We were in the room next to Nap North, and immediately after the test we learned that we had beaten them by 15 pts ... we thought we were in pretty good shape until this massive truck named IMSA rolled us over.

W/e though ... OPRF had its best placement ever, and probably among the best we will ever have.  We could have been 8th, too bad I didn't have the right orals materials.

JB

Heh, I only qualified for 3AA calculator team and took 8th, which is good for a non-math powerhouse. Next year though, mark my words! I was never any good with geometry anyways. :P

[quote="besttate"]First place Freshman 4AA written! :) NT only got 4th though... but we can't really compete with Naperville or IMSA[/quote]

Wow! That's impressive! It's really a huge loss for our state's quadratic-equation-solving talent that you're going to Andover. :(

Next year, Naperville Central's gonna own majorly, for the first time in a long while. Mark my words.


hi everyone.

i, after joining many tournaments, have decided to make my own! it will be a single elimination, countdown (best of three) tournament. Seeding is done based on random.org.

to join, please write that you would like to join here AND private message me. one is fine, but i'd like both. i will allow up to 16 members (although i doubt it'll even get that high).

SIGN UPS END DECEMBER 20TH. ask if you have questions.  :lol:

If this is a FTW tournement, please consider the FTW other stuff forum before posting here.

SIGN UPS:
athunder
ernie
theprodigy
007math
SuperMathBoy
someperson01

NEED TWO MORE!

Only 8 people?  I hope those 8 are active users...

I'll join, if this isn't a FTW tourney.

Eh, otherwise I'll still join

SIGN UPS:
athunder
ernie
theprodigy
007math
SuperMathBoy
someperson01
Twin Prime Conjecture
BOGTRO

EVERYONE MUST SEND ME A PM IN THE NEXT TWENTY FOUR HOURS TO PROVE THAT YOU ARE ACTIVE USERS! you can still sign up, you will be on the waiting list though.

active:

athunder
007math
SuperMathBoy
Twin Prime Conjecture
someperson01
BOGTRO

waiting on:
xpmath
ernie

!YOU HAVE ONE MORE DAY OR ELSE YOU'RE ELIMINATED!

I...don't think I signed up for this? Take me out.

Countdown isn't really my sort of competition.

active and confirmed:

athunder
007math
SuperMathBoy
Twin Prime Conjecture
someperson01
BOGTRO
AIME15

NEED ONE MORE.

ill join..havent joined for long but im pretty active ill pm u too i guess

Request to make it best of 5.  Lucky memmers can win 2 of 3.

BEST OF FIVE:

seeding:

1. 007math
2. Twin Prime Conjecture
3. AIME15
4. athunder
5. ReesesPieces
6. someperson01
7. BOGTRO
8. SuperMathBoy

1. 007math vs. 8. SuperMathBoy (RESULTS: )
2. Twin Prime Conjecture vs. 7. BOGTRO (RESULTS: )
3. AIME15 vs. 6. someperson01 (RESULTS: )
4. athunder vs. 5. ReesesPieces (RESULTS: )

PLEASE HAVE BOTH OPPONENTS PM ME THE RESULTS. HAVE ALL RESULTS IN BY DECEMBER 23RD (or random.org will determine who is out.)
good luck!

1. 007math
2. Twin Prime Conjecture
3. AIME15
4. athunder
5. ReesesPieces
6. someperson01
7. BOGTRO
8. SuperMathBoy

1. 007math vs. 8. SuperMathBoy (RESULTS: )
2. Twin Prime Conjecture vs. 7. BOGTRO (RESULTS: )
3. AIME15 vs. 6. someperson01 (RESULTS: AIME15 def someperson01 3-0)
4. athunder vs. 5. ReesesPieces (RESULTS: )

AIME15 is the champion because AIME15 was the only one to win a game.  YAY!

athunder vs. aime for championship...
no one actually followed through; hence, neither did i.


number theory
greatest common divisor
prime factorization

So some friends and I noticed this afternoon that the prime factorization of [tex]2^{5n}-1[/tex], or in general, [tex]a^{5n}-1[/tex] has way too many factors that end in 1.  It's usually all of them except for a few trivial ones in the beginning.  Look at some, you'll see what I mean.

Any ideas why?

And, if you want a real problem (simple), see if you can show why this trait would also imply it for [tex]a^{5n}+1[/tex].

If 5n = p-1, and gcd (a, 5n+1) = 1, then by Fermat's Little Theorem, a^5n :equiv: 1 (mod 5n+1)

A number of such primes are quite small: n = 2, p = 11; n = 6, p = 31; n=8, p = 41; n = 12, p = 61; n = 14, p = 71, etc., so a^5n will have at least one factor that ends in one.  Moreover,

As an aside, in general, what values of n will make 5n+1 prime?  Clearly, n must be even.  Also, if n :equiv: 4 (mod 6), 5n+1 will be a multiple of 3.  Any other ideas?

Well [tex]a^{10n}-1=(a^{5n}+1)(a^{5n}-1)[/tex] to answer your second question.


Let $ x_1,...,x_n$ be positive real numbers. Prove that:

$ \displaystyle\sum_{k\equal{}1}^{n}x_k\plus{}\sqrt{\displaystyle\sum_{k\equal{}1}^{n}x_k^2} \le \frac{n\plus{}\sqrt{n}}{n^2} \left( \displaystyle\sum_{k\equal{}1}^{n} \frac{1}{x_k} \right) \left( \displaystyle\sum_{k\equal{}1}^{n} x_k^2 \right).$

[quote="moldovan"]Let $ x_1,...,x_n$ be positive real numbers. Prove that:

$ \displaystyle\sum_{k \equal{} 1}^{n}x_k \plus{} \sqrt {\displaystyle\sum_{k \equal{} 1}^{n}x_k^2} \le \frac {n \plus{} \sqrt {n}}{n^2} \left( \displaystyle\sum_{k \equal{} 1}^{n} \frac {1}{x_k} \right) \left( \displaystyle\sum_{k \equal{} 1}^{n} x_k^2 \right).$[/quote]
It's two obvious inequalities.  :wink:


Prove that $a,b,c\geq 0$ then $\sum{\sqrt{a^4+b^4/2+c^4/2}}\geq \sum{\sqrt{a^4+b^3c}}$


In a $3\times 3$ square , fill in $9$ distinct natural number , such that the product of three numbers in every row and every column are the same . We obtain $6$ similar product which we call it $P$ . Find the least value of $P$ .

 :)

[quote="Kondr"]Using numbers of the form $2^a3^b$ where $0\leq a,b \leq 2$ we get $P=2^3 3^3=216$[/quote]

I believe thats no correct  :) And it can be less than $216$ .

[quote="a_vakilian"]if we consider $r_i/c_i $ product of row/column $i$ then:
$r_1r_2r_3c_1c_2c_3=(a)^2$ which $a$ is the product of those $9$ numbers. and we knwo that $r_1,r_2,r_3,c_1,c_2,c_3$ are the same so :
$P^6=a^2 ,min(a)=9! \Rightarrow P^6 \geq 9!^2$.
tell me if there is wrong with me![/quote]

I can see that you are using number from $1$ to $9$ to fix into the square but the question is , the number $1$ to $9$ dont nessecary fullfill the condition for all $r_i=c_j$ .  Notice that if $7$ is in one of the row (say row 1) then it follows that the second and third row must contain $7$ which cannot happen in your case since there are only one $7$ .

For 216 it's
1 6 36
18 4 3
12 9 2
The idea of exponents of 2 and 3 is verz simple, I believe I don't have to coment it. Using just one prime leads to product $2^{15}$, but i don't know how is it with using more primes. If u can show a solution with product lower then 216, please post it.

[quote="Kondr"]For 216 it's
1 6 36
18 4 3
12 9 2
The idea of exponents of 2 and 3 is verz simple, I believe I don't have to coment it. Using just one prime leads to product $2^45$, but i don't know how is it with using more primes. If u can show a solution with product lower then 216, please post it.[/quote]

Sorry I also dont know how to prove it ( or else it would be in the "solved" section :P )

But there is a way to make the product $120$  which is not hard to verify ...  :roll:

So anyone else has any idea ?

If there were 4 or more distinct primes that divide the numbers, all of them will have to appear in one row and the product of the row will be greater or equal to 2.3.5.7=210. If there were only two (which I wrongly supposed to be the best), the product will be greater or equal to 216 and when there was only one, the product will be greater or equal to $2^{15}$. So there must be exactly three distinct primes and all the numbers must be of the form $2^x.3^y.5^z$. Let $a$, $b$ and $c$ be the number of diferent values of $x$, $y$ and $z$. Because there must be 9 diferent number in the square,
$a.b.c \geq 9$.  Because $a\geq b\geq c$, $a\geq 3$.  Sums of x, y and z must be in each row and column the same. Because x must have 3 diferent values, these  can be at least 0,1 and 2, so the sum of x in a row is at least 2. Because y and z must be for some numbers in the square positive (else there  were only 2 distinct primes, which is discussed above), sumd of y and z are both at least 1.  If the sum of x in a row was only two, the matrice of x in the square should look like this
0 0 2
1 1 0
1 1 0
after a few tries with posible values of y and z it's obvious that there is no solution for $\sum_{x \in {row}}{x}= 2$. Because of that the sum must be at least 3,
$P=2^{\sum_{x \in {row}}{x}}.3^{\sum_{y \in {row}}{y}}.5^{\sum_{z \in {row}}{z}}\geq 2^3.3^1.5^1=120$. Because for P=120 there is a solution, the task is (maybe  :huh: ) solved.
  3    2   20
 10  12   1
   4    5   6


Solve the equation: $ 3^{x}\plus{}2*4^{x}\plus{}6^{x}\equal{}|x|^{3}$

[b]Sketch:[/b]

Over negative x, the LHS rises strictly from 0 to 4 and the RHS strictly falls from indefinitely positive to 0, so there is 1 solution here. This cannot be solved algebraically, but by inspection, -1 works.

Over positive x, both are increasing. If one wishes to avoid calculus, one can easily show that for the interval [0,1], the LHS is greater than the RHS, and show that finite differences spaced by x-interval length 1 are larger for the LHS than the RHS, and use induction, to show that LHS > RHS.

Hence, the only solution is [b]x=-1[/b]


linear algebra
matrix
modular arithmetic

Find the number of matrices $ (m\times n)$, such that each cell contains either $ 1$ or $ 0$, and the sum of numbers in each column and row is odd.

Isn't it infinite? Because take an $ n\times n$ matrix and make one diagonal all 1s and make all the other cells 0.

The number of matrices is at least limited by number $ 2^{mn}$, so it can not be infinite.

-Well, $ m,n$ are fixed, if that was not obvious.

[hide]If $ m \not\equiv n \pmod{2}$, then the sum of all the rows and the sum of all the columns have opposite parity, so $ 0$.

If $ m \equiv n \pmod{2}$, arbitrarily fill an $ m \minus{} 1 \times n \minus{} 1$ subgrid. For each row and column with one space left, fill the last space with whatever is needed. After that there is one corner space left; fill it with whatever. Because the sum of all rows and all columns must have the same parity, this final corner space must be able to satisfy both the row and column it is in. So $ 2^{(m \minus{} 1)(n \minus{} 1)}$ is the answer here.[/hide]


algebra
polynomial
modular arithmetic
number theory
least common multiple
greatest common divisor
induction

[b]Step 1:[/b] Suppose $P$ has degree $d$. Let $Q$ be the polynomial of degree at most $d$ with $Q(x)=q_x$ for $0\leq x\leq d$. Since the $q_x$ are all integers, $Q$ has rational coefficients, and there exists $k$ so that $kQ$ has integer coefficients. Then $m-n|kQ(m)-kQ(n)$ for all $m,n\in \mathbb N_0$.

[b]Step 2[/b] We show that $Q$ is the desired polynomial.

Let $x>n$ be given. Now
\[kq_x\equiv kq_m\pmod{x-m}\text{ for all integers }m\in[0,d]\]
Since $kQ(x)$ satisfies these relations as well, and $kq_m=kQ(m)$,
\[kq_x\equiv kQ(x)\pmod{x-m}\text{ for all integers }m\in[0,d]\]
and hence
\[kq_x\equiv kQ(x)\pmod{\text{lcm}(x,x-1,\ldots, x-d)}. \;(1)
\]
Now
\begin{align*}
\text{lcm}(x,x-1,\ldots, x-i-1)&=\text{lcm}[\text{lcm}(x,x-1,\ldots, x-i),x-i-1]\\
&=\frac{\text{lcm}(x,x-1,\ldots, x-i)(x-i-1)}{\text{gcd}[\text{lcm}(x,x-1,\ldots, x-i),(x-i-1)]}\\
&\geq \frac{\text{lcm}(x,x-1,\ldots, x-i)(x-i-1)}{\text{gcd}[x(x-1)\cdots(x-i),(x-i-1)]}\\
&\geq \frac{\text{lcm}(x,x-1,\ldots, x-i)(x-i-1)}{(i+1)!}
\end{align*}
so by induction $\text{lcm}(x,x-1,\ldots, x-d)\geq \frac{x(x-1)\cdots (x-d)}{d!(d-1)!\cdots 1!}$. Since $P(x), Q(x)$ have degree $d$, for large enough $x$ (say $x>L$) we have $\left|Q(x)\pm\frac{x(x-1)\cdots (x-d)}{kd!(d-1)!\cdots 1!}\right|>P(x)$. By (1) $kq_x$ must differ by a multiple of $\text{lcm}(x,x-1,\ldots, x-d)$ from $kQ(x)$; hence $q_x$ must differ by a multiple of $\frac{x(x-1)\cdots (x-d)}{kd!(d-1)!\cdots 1!}$ from $Q(x)$, and for $x>L$ we must have $q_x=Q(x)$.

Now for any $y$ we have $kQ(y)\equiv kQ(x)\equiv kq_x \equiv kq_y\pmod{x-y}$ for any $x>L$. Since $x-y$ can be arbitrarily large, we must have $Q(y)=q_y$, as needed.

Nice!

We consider the polynomial $f$ of the minimal degree such that $a_n \le f(n)$ for all but finitely many $n$. Let $d$ be the degree of $f$. We consider a polynomial $g$ of degree $d$ such that $g(i)=a_i$ for $1 \le i \le d+1$. By the Lagrange Interpolation formula, this polynomial is of rational coefficients and uniquely defined. Let $N>0$ be a constant such that $h(x)=Ng(x)$ is a polynomial with integer coefficients. We replace the sequence $a_n$ by $b_n=Na_n$. This does not affect our divisibility relation and the other conditions are scaled accordingly. Choose $m$ sufficiently large, in particular $m>c$ where $c>0$ is a constant we shall disclose later in the proof. Now, we have \begin{align*} m-i \mid (b_m-b_i)-(h(m)-h(i)) = (b_m-h(m)) \end{align*} where $1 \le i \le d+1$. Therefore, we have the relation: \begin{align*} \operatorname{lcm}(m-1,\dots, m-d, m-d-1) \mid |b_m-h(m)| \end{align*} and we observe that $|b_m-h(m)|_{m \ge 1}$ is bounded above by a polynomial of degree $d$, i.e., there exists a constant $a>0$ such that $|b_m-h(m)| \le am^d$. However, we see that the relation \begin{align*} x_1\cdot \dots \cdot x_k \mid \operatorname{lcm}(x_1,\dots, x_k)\cdot \Pi \gcd (x_i,x_j) \end{align*} holds for all integers $x_1,\dots, x_k$, where in the last product the gcd is taken over all pairs $1 \le i < j \le k$. This result is established by comparing the $v_p$'s on each side.  Now, this directly translates into an inequality $A \le B$ if $A \mid B$. Notice that $\gcd (a,b) \le |a-b|$. Therefore, we have \begin{align*} am^d \ge \operatorname{lcm}(m-1,...,m-d-1) \ge \frac{(m-1)\cdot \dots \cdot (m-d-1)}{\Pi_{1 \le i<j \le d+1} \gcd(m-i,m-j)} \ge \frac{(m-1)\cdot \dots \cdot (m-d-1)}{\Pi_{1 \le i < j \le d+1} (j-i)}=O(m^{d+1}), \end{align*} yielding a contradiction for all sufficiently large $m$, unless $b_m=h(m)$. We deduce that $b_m=h(m)$ for all sufficiently large $m$, i.e., for all $m>M$ for some constant $M>0$. Now, we take $n \le M$ and notice that $m-n \mid b_m-b_n$ implies that $m-n \mid |b_n-h(n)|$ which further yields, due to the size of $m$, that $b_n=h(n)$.

In summary, we conclude from the above deductions that $b_n=h(n)$ for all $n \in \mathbb{N}$ and so, $a_n=g(n)$ for all $n \in \mathbb{N}$. The result follows.

[b]Comment[/b] The result is sharp, namely, we cannot reduce it to showing that $g \in \mathbb{Z}[X]$. A counter example can be the sequence $a_n=\binom{n}{2}+2016$ which clearly, satisfies all our conditions but is not in $\mathbb{Z}[X]$.

Let $d=\deg P$. Let $Q(x)$ be the unique polynomial with degree at most $d$ such that $Q(x)=q_x$ for all $0\le x\le d$. It exists because we can set up a system of $d+1$ variables and linear equations, and the solution exists and is unique because the Vandermonde determinant is nonzero. This also shows that $H$ has rational coefficients. Pick $k\in\mathbb{Z}$ such that $H=kQ\in\mathbb{Z}[x]$, and let $a_n=kq_n$. It now suffices to show that $a_n=H(n)$ for all $n\ge 0$, where we know that $H(n)=n$ for all $0\le n\le d$, $\deg H\le d$, and $m-n\mid a_m-a_n$ for all $m\not=n$.

Suppose $n\ge d+1$. Then, we have that
\begin{align*}
a_n &\equiv a_0 = H(0)\pmod{n} \\
a_n &\equiv a_1 = H(1)\pmod{n-1} \\
&\vdots \\
a_n &\equiv a_d = H(d) \pmod{n-d}.
\end{align*}
By the Chinese Remainder Theorem, this has a unique solution $\pmod{f(n)}$ where $f(n):=\mathrm{lcm}(n-d,n-d+1,\ldots,n-1,n)$. Note that one solution that definitely works is $H(n)$ since $H\in\mathbb{Z}[x]$. Thus, $a_n\equiv H(n)\pmod{f(n)}$. The key point is that $n^{d+1}/f(n)$ is bounded above by some constant that only depends on $n$ (I think something like $d\cdot (d-1)^2\cdots 2^{d-1}$ works), so in particular, $f(n)$ grows faster than $P(n)$. Eventually then, we must have $f(n)>100k P(n)$, and since $|a_n|<k P(n)$ and $a_n\equiv H(n)\pmod{f(n)}$, we see that $a_n=H(n)$ for sufficiently large $n$. So $a_n=H(n)$ for all $n\ge N$.

It remains to show that $a_n=H(n)$ for all $n<N$. We see that $a_n\equiv a_m=H(m)\equiv H(n)\pmod{m-n}$ for all $m\ge N$, so by taking $m$ to be sufficiently large, we learn that $a_n=H(n)$, as desired. Dividing by $k$, we get that $q_n=Q(n)$ for all $n$.

For those interested, [url="https://arxiv.org/abs/2005.08833"] here[/url] is a massive generalization (posted today!)

Let \(d\) be the degree of \(P\), and let \(Q\) be the (unique) rational polynomial with \(Q(i)=q_i\) for \(i=0,\ldots,d\). Appropriately scale up everything so that \(Q\) has integer coefficients.

[color=red][b]Lemma:[/b][/color]      For all \(d\),     \[\operatorname{lcm}(n,n-1,\ldots,n-d)=O\big(n^{d+1}\big).\]

[i]Proof.[/i]      Evidently \(\operatorname{lcm}(n,n-1,\ldots,n-d)\le n(n-1)\cdots(n-d)=O\big(n^{d+1}\big)\). The lower bound can be attained easily via basically any bound; for instance,     \[\operatorname{lcm}(n,\ldots,n-d)\ge\frac{n(n-1)\cdots(n-d)}{\prod_{i<j}\gcd(n-i,n-j)}\ge\frac{n(n-1)\cdots(n-d)}{\prod_{i<j}(j-i)}=O\big(n^{d+1}\big).\] \(\blacksquare\)

[color=red][b]Claim:[/b][/color]      For sufficiently large \(n\), we have \(q_n=Q(n)\).

[i]Proof.[/i]      We must have \(q_n\equiv q_i=Q(i)\equiv Q(n)\pmod{n-i}\) for \(i=0,\ldots,d\), so     \[q_n\equiv Q(n)\pmod{\operatorname{lcm}(n,\ldots,n-d)}.\]     If \(q_n\ne Q(n)\), then for some nonzero \(k\) we have \(q_n=Q(n)+k\operatorname{lcm}(n,\ldots,n-d)=O\big(n^{d+1}\big)\), which will exceed \(P(n)\) in absolute value for sufficiently large \(n\). \(\blacksquare\)

To finish the proof, observe that for every \(m\), we have \(q_m\equiv q_n=Q(n)\equiv Q(m)\pmod{n-m}\) for sufficiently large \(n\). By taking \(n\) large, \(q_m=Q(m)\), as needed.

Let $P$ have degree $k$. Let the polynomial $Q$ of minimal degree such that $Q(i)=q_i$ for each $0\le i\le k$ be $Q^*$. Note that $Q^*$ is a rational polynomial with degree at most $k$. Let $d\in\mathbb{Z}_{>0}$ be minimal with $dQ^*\in\mathbb{Z}[x]$. Define $p_i=dq_i$ for each $i$ and note the $p_i$ satisfy the same properties as the $q_i$ by considering $P'=dP(n)$, and moreover the minimal polynomial of $p_0,\dots,p_k$ is an integer polynomial $\tilde{Q}$. Now, consider $n>k$. Observe that we must have $\tilde{Q}(n)\equiv p_n\pmod{n-i}$ for each $0\le i\le k$ because of the condition and the fact that $\tilde{Q}$ is an integer polynomial. Note that any $n-i,n-j$ have common divisor dividing $k!$, hence we can write
\[n\cdot \binom{n}{k} \mid \tilde{Q}(n)-p_n.\]
For large values of $n$, $|\tilde{Q}(n)+ an\cdot \binom{n}{k}|>10|P(n)|$ for all $a\ne0$ so $p_n=\tilde{Q}(n)$. Now, use large $n$ to uniquely determine every value of $p_i$ for which the previous argument does not work, and note that $p_i=\tilde{Q}(i)$ is the only feasible solution $p_i$ to the modular equations we get. Hence, $q_n=Q^*(n)$ for each $n$, as desired.

[quote=MithsApprentice]Suppose $\, q_{0}, \, q_{1}, \, q_{2}, \ldots \; \,$ is an infinite sequence of integers satisfying the following two conditions:

(i)  $\, m-n \,$ divides $\, q_{m}-q_{n}\,$ for $\, m > n \geq 0,$
(ii) there is a polynomial $\, P \,$ such that $\, |q_{n}| < P(n) \,$ for all $\, n$

Prove that there is a polynomial $\, Q \,$ such that $\, q_{n}= Q(n) \,$ for all $\, n$.[/quote]

Suppose $\text{deg} \ P = d$. Then let by Lagrange Interpolation formula, there exists $Q \in \mathbb{Q}[x]$ of degree $d$ such that $Q(i) = q_i$ for $0 \le i \le d$. For now, we alter $q_i \to kq_i$, $P \to kP = P'$ and $Q \to kQ = Q'$ such that $Q' \in \mathbb{Z}[x]$, and the problem is still true.

We wll know prove that $Q'$ satisfies the requirement of the problem.
Fix an integer $\ell > d$. Notice that we have
\[ \ell - i \mid kq_{\ell} - kq_i, \ \forall 0 \le i \le d \]
and since $Q'(i) = kq_i$ for $0 \le i \le d$, we have $\ell - i \mid kq_{\ell} - Q'(i)$. But since $Q' \in \mathbb{Z}[x]$, we have $\ell - i \mid Q'(\ell) - Q'(i)$, which forces
\[ \ell - i \mid kq_{\ell} - Q'(\ell) \ \text{for } 0 \le i \le d \]
This forces
\[ \text{lcm} ( \ell, \ell - 1, \dots, \ell - d) \mid kq_{\ell} - Q'(\ell) \]
[b][color=#f00]Claim.[/color][/b] $\text{lcm}(\ell, \ell - 1, \dots, \ell - d) = O(\ell^{d + 1})$.
[i]Proof.[/i] We know that $\text{lcm}(\ell, \ell - 1, \dots, \ell - d) \le \ell(\ell - 1) \dots (\ell - d) = O(\ell^{d + 1})$, and the lower bound is attained by
\[ \text{lcm}(\ell, \dots, \ell - d) \ge \frac{\ell(\ell - 1)\dots (\ell - d)}{\prod_{i < j} \gcd(\ell - i, \ell - j)} \ge \frac{\ell(\ell - 1) \dots (\ell - d)}{\prod_{i <j} \gcd(i,j)} = O(\ell^{d + 1}) \]
------------------
[b][color=#f00]Claim.[/color][/b] $Q'(n) = kq_n$ for all sufficiently large integer $n$.
[i]Proof.[/i]  From the condition of the problem, we have $|kq_n| < P'(n)$ for all $n \in \mathbb{N}$.
So, $kq_{\ell} - Q'(\ell) \le |Q'(\ell)| + |P'(\ell)|$, and therefore for sufficiently large $n$, if $Q'(n) \not= kq_n$, we must have
\[ n(n - 1) \dots (n - d) \le |kq_n - Q'(n)| \le |P'(n)| + |Q'(n)| \]
which is a contradiction as $P' + Q'$ is of degree $d$.
------------
[b][color=#f00]Claim.[/color][/b] $Q'(n) = kq_n$ for all positive integers $n$.
[i]Proof.[/i] Suppose there exists a positive integer $\ell$ such that $Q'(\ell) \not= kq_{\ell}$. We have
\[ n - {\ell} \mid Q'(n) - Q'(\ell) \ \text{and} \ n - \ell \mid kq_n - kq_{\ell} \]
for all sufficiently large $n$. Since $kq_n = Q'(n)$ for all sufficiently large $n$, then we must have
\[ n - {\ell} \mid Q'(\ell) - kq_{\ell} \]
for all sufficiently large $n$. However, this is a contradiction, as we can take $n - \ell > |Q'(\ell) - kq_{\ell}| + 10^{100}$, and we are done.

To conclude, we have $kQ(n) = Q'(n) = kq_n$, which forces $Q(n) = q_n$ for all $n \in \mathbb{N}$.

Let $P(x)$ have degree $k$. Suppose that $Q(x)$ is the polynomial of degree $k$ such that $Q(x)=q_x$ for $1\leq{}x\leq{}k$. We now claim that if we fix the first $k$ terms of the sequence $q_i$ such that it satisfies the conditions, the rest of the sequence is fixed. (If it is fixed, it also must be equivalent to the polynomial $Q(x)$, since $Q(x)$ satisfies the conditions). \\

Notice that $Q$ must be a polynomial with rational coefficients (comes from finite differences), therefore there is an integer $c$ such that $cQ$ is a polynomial with integer coefficients. For some very large $x>n$, we find that
\[\text{lcm}{}(x,x-1,\dots{},x-k)\geq{}\frac{x(x-1)\dots{}(x-k)}{ck!(k-1)!\dots{}1!}\]
and by bounding on $q_x$ (Since by the second condition, we must have that $|q_x|<P(x)$), we find that we must have $Q(x)=q_x$ for all $x$, and we are done.

The hardest part of the problem is really getting started.

Fix $\deg Q = d = \deg P$, and let $Q$ be the polynomial with degree $d$ that agrees with $q_i$ for $0 \leq i \leq d$. We can assume that $Q$ is an integer polynomial; otherwise just transform $Q \to kQ$ where $kQ \in \mathbb Z[X]$ and show that $kQ(n) = kq_n$ for every $n$.

Now, for any $n \geq d$, note that $$q_n \equiv Q(n) \pmod {\operatorname{lcm}(n, n-1, \dots, n-d)}.$$ Call this $A_n$.

[b][color=#f00]Claim.[/b][/color] $A_n = O(n^{d+1})$.

[i]Proof.[/i] Intuitively, this is because for large $n$, the numbers ``don't share many common factors." More rigorously, $$A_n > \frac{n(n-1)\dots(n-d)}{\prod_{i<j} \gcd(i, j)} = O(n^{d+1}). \ \blacksquare$$

To finish, pick very large $n$ and suppose that $q_n \neq Q(n)$; then it follows that $|q_n - Q(n)| > A_n$. But by the claim this implies $|q_n| > P(n)$, contradiction! Thus, $q_n$ and $Q(n)$ agree at infinitely many values, and $Q(n) = q_n$ for all $n$.

Let $N$ be a sufficiently large integer and let $Q$ be the interpolation polynomial of the points $(0, q_0), (1, q_1), \dots, (N, q_N)$. We'll show that $Q$ is the desired polynomial. We can assume that $Q$ is an integer polynomial; otherwise just transform $Q \to kQ$ where $kQ \in \mathbb Z[X]$ and show that $kQ(n) = kq_n$ for every $n$. We'll do it by induction. Base  case is clear. For induction step, assume $Q(n) \neq q_n$ for some $n$. Since $q_n \equiv q_i = Q(i) \equiv Q(n) (n - i)$, so $\text{lcm}(1, 2, \dots, n-1) \mid q_n - Q(n)$. Therefore there exists $k$ such that $q_n = Q(n) + \text{lcm}(1, 2, \dots, n-1)k$. Now consider the following claim:

[b][color=red]Claim:[/color][/b] $\text{lcm}(1, 2, \dots, n)$ grows faster than any polynomial.

[i]Proof.[/i] Assume not. Then there exists $k$ such that $n^k > \text{lcm}(1, 2, \dots n)$ for all sufficiently large integer $n$. But we have $\text{lcm}(1, 2, \dots, n) > \prod_{p \le n} p^{\log_p(n) - 1} = \prod_{p \le n} \frac{n}{p} = \frac{n^{\pi(n)}}{p_1p_2\cdots p_{\pi(n)}} > \frac{n^{k+1}}{p_1p_2\cdots p_{k+1}}$, hence we get $n^k > \frac{n^{k+1}}{p_1p_2\cdots p_{k+1}}$ for all sufficiently large $n$, a contradiction. $\blacksquare$

Note that $|Q(i)| < P(i)$ for all $0 \le i \le N$, so $\deg Q \le \deg P$. Thus $q_n$ has to be equal to $Q(n)$, otherwise we get $\text{lcm}(1, 2, \dots, n)$ is bounded by some polynomial, contradicting the claim. Hence the induction step is proved. $\blacksquare$

[b]Remark.[/b] In fact, we have $\text{lcm}(1, 2,  \dots, n) = n \cdot \text{lcm}(\binom{n-1}{0}, \binom{n-1}{1}, \dots, \binom{n-1}{n-1}) \ge \binom{n-1}{0} + \binom{n-1}{1} + \dots + \binom{n-1}{n-1} = 2^{n-1}$, which immediately implies the claim.

The main idea of this problem is the following lemma:

Lemma. For all non-negative integers $d$, we have
\[\text{lcm}(n,n-1,\ldots,n-d)=O(n^{d+1})\]
Proof. For the upper bound, simply note that
\[\text{lcm}(n,n-1,\ldots,n-d)\leq n(n-1)\cdots(n-d)=O(n^{d+1}),\]
while for the lower bound, we use the fact that for $1\leq i < j\leq d$, we have $\gcd(n-i, n-j)\leq j-i$, which means
\[\text{lcm}(n,n-1,\ldots,n-d)\geq \frac{n(n-1)\cdots(n-d)}{\prod_{1\leq i < j\leq n}(j-i)}=O(n^{d+1}),\]
so the lemma is true. $\blacksquare$

Let $d$ be the degree of $P$. So, $q_n=o(P(n))=o(n^d)$ for sufficiently large $n$.
By Lagrange Interpolation, there exists a unique rational polynomial $R$ of degree at most $d$ such that $R(i)=q_i$ for each $0\leq i\leq d$.

Claim: $R(n)=q_n$ for all sufficiently large $n$.
Proof. Note that we have
\[q_n\equiv q_i\equiv R(i)\equiv R(n)\pmod{n-i}\]
for each $0\leq i\leq d$. Hence, we have
\[q_n\equiv R(n)\pmod{\text{lcm}(n-1,n-2,\ldots,n-d)}\Longrightarrow q_n=k\ \text{lcm}(n-1,n-2,\ldots,n-d)+R(n)\]
So, if $k\neq0$, then $q_n=O(n^{d+1})$. But we also have $q_n=o(n^d)$ for sufficiently large $n$. This means $O(n^{d+1})=o(n^d)$, which is absurd. Hence, $k=0$ is forced for large $n$, which means $q_n=R(n)$, as desired. $\blacksquare$

Finish:
Note that for any $m$ and sufficiently large $n$, we have that
\[q_m\equiv q_n\equiv R(n)\equiv R(m)\pmod{n-m}\]
So, $n-m\mid q_m-R(m)$ for all sufficiently large $n$, which forces $q_m=R(m)$.

So, $R$ is the required construction for $Q$. $\blacksquare$


Find the resultant of the two concurrent forces F=32kN, which makes an angle -120 degrees with respect to the positive x-axis, and G=72kN, which makes an angle 45 degrees with respect to the positive x-axis.

For $ F$, instead of thinking of it as a $ \minus{}120^{/circ}$ angle, think of it as a $ 30^{\circ}$ angle with respect to the negative y.  The the $ x$ component of $ F$ is given by $ 32,000\sin{30^{\circ}}\equal{}16,000$ in the negative direction.  Then by Pythagoras, $ y\equal{}\sqrt{32000^{2}\minus{}16000^{2}}\equal{}27712.8$ so $ F\equal{}(\minus{}16000,\minus{}27712.8)$.  For $ G$, $ x\equal{}y\rightarrow 2x^{2}\equal{}72000\rightarrow x\equal{}y\equal{}6000$.  So $ G\equal{}(6000,6000)$.  Then just add the two so $ R\equal{}(\minus{}10000,\minus{}21712.8)$.  So those are your components.  Then use Pythagoras to get the total length.


Έστω $ p$ ένας πρώτος αριθμός .Αν $ u\geq 2$ και $ \frac{b^n\minus{}1}{b\minus{}1}\equal{}p^u$ τότε δείξτε ότι ο n είναι πρώτος .

[quote="silouan"]Έστω $ p$ ένας πρώτος αριθμός .Αν $ u\geq 2$ και $ \frac {b^n \minus{} 1}{b \minus{} 1} \equal{} p^u$ τότε δείξτε ότι ο n είναι πρώτος .[/quote]

συμβολίζω f το πλήθος των μικρότερων σχετικών πρώτων με τον αριθμό

$ b^{f(n) \plus{} 1} \equiv b (mod n) \Leftrightarrow p^u(b^{f(n) \plus{} 1} \minus{} 1) \equiv (b \minus{} 1)p^u (mod n) \Leftrightarrow$
$ p^u(b^{f(n) \plus{} 1} \minus{} 1) \equiv (b^n \minus{} 1) (mod n)$
έχουμε τώρα:
$ p^u \equiv 1 (mod n)$ (δεν είμαι απολύτως σίγουρος ότι αυτό ισχύει πάντα)
$ \Rightarrow kb^{f(n) \plus{} 1} \equiv b^n (mod n) \Leftrightarrow$
$ n \equal{} f(n) \plus{} 1 \le f(n) \plus{} 1 \Leftrightarrow$
$ n \minus{} 1 \equal{} f(n) \Leftrightarrow n \equal{}$ πρώτος

[quote="maniopas"][quote="silouan"]Έστω $ p$ ένας πρώτος αριθμός .Αν $ u\geq 2$ και $ \frac {b^n \minus{} 1}{b \minus{} 1} \equal{} p^u$ τότε δείξτε ότι ο n είναι πρώτος .[/quote]
$ p^u \equiv 1 (mod n)$ (δεν είμαι απολύτως σίγουρος ότι αυτό ισχύει πάντα)
[/quote]

Ωραία η προσέγγιση για να αποδείξεις το ζητούμενο αλλά αυτό που γράφεις εδώ νομίζω δεν ισχύει πάντα αλλά και να ισχύει, πρέπει σίγουρα να δώσεις μια δικιολόγηση αφού δεν είναι κάτι "φανερό" :huh:


AMC
AIME
geometry
ratio
area of a triangle
Heron\u0027s formula

No solution on Kalva. Could somebody solve this please?

13. The triangle ABC has AB = 9 and BC/CA = 40/41. What is the largest possible area for ABC?

[quote="beta"][hide]Label BC=40x, AC=41x.

Heron's formula on ABC

area $= \frac{1}{4} \sqrt{(81x^2-1)(81^2-81x^2)}$

Let $81x^2-1=a, 81^2-81x^2=b$, we know that a, b are both positive reals

By AM-GM

$\frac{a+b}{2} = 3280 \geq \sqrt{ab}$

so $\sqrt{ab}$= 1/4 of desired area which maximize at 3280

area maximize at (1/4)(3280) = 820[/hide][/quote]

Good. That's what I got. Using the same method, too. I just wanted to make sure I didn't do anything dumb as usual.

It is well known that for fixed points A and B and a ratio p, the locus of points C so that BC/CA=p is a circle, called the Circle of Apollonius (or a line if p=1). Since the area of our triangle is 1/2*b*h=9/2*d, where d is the distance from C to line AB, we must maximize d. This maximum distance is the radius of our circle of appolonius; we just need to find this radius.

We can do this by finding two diametrically opposite points on the circle. First, there is a point C1 between A and B so that BC1=40x and C1A=41x, which means AB=81x=9 and x=1/9, so C1B=40/9. There is also a point C2 with B between C2 and A so that BC2=40y and AC2=41y, meaning AB=y=9 and BC2=360. C1C2 is a diameter of the circle, so the radius is 1/2(360+40/9)=1640/9, and our answer is 9/2*1640/9=820.

Wow. That's a really cool solution. I would've never thought of it though =P.

[quote="litoxe2718281828"]I set up A(0,0), B(9,0), and P(x,y), and
\[\frac{\sqrt{x^2+y^2}}{\sqrt{(x-9)^2+y^2}}=\frac{41}{40}\]
It seems to me like clearing denominator, squaring both sides and simplifying would get you a circle, and since there's no $y^1$ term the center is on the x-axis and the height of the triangle is just the radius of this circle, which is easy to find...

But it's not working! Can someone help?[/quote]

You have two separate equations, right?
\[\sqrt{x^2+y^2}=41\]
and
\[\sqrt{(x-9)^2+y^2} =40\]
[hide]
so if you square the first one and square the second one then you get:
\[\ x^2+y^2=41^2\] and,
\[\ (x-9)^2+y^2=40^2\] or,
  \[\ (x^2-18x+81)+y^2=41^2\]
so by subtracting the third equation from that last one,
\[\ -18x+81=41^2-40^2  \]
\[\ -18x+81=(41-40)(41+40) \]
\[\ -18x = 0 \] is that right??
If x = 0 then we have that \[\ y = 41 \]

 [/hide]

[quote]How do you get that area equation from Heron's formula?[/quote]

You have to jiggle some terms around after applying Heron's formula.

$a=9, b=40x, c=41x, s=\frac{81x+9}{2}$,

then

\[\sqrt{s(s-a)(s-b)(s-c)} = \frac{\sqrt{(81x+9)(81x-9)(x+9)(-x+9)}}{4} = \frac{\sqrt{(81^2 x^2 - 9^2)(81-x^2)}}{4}\]

Now under the radical, remove a factor of 81 from the first factor and put it in the second factor, and we get beta's formula.

(I just realized, I've labelled the sides differently from the notation in the problem, but hopefully it is clear.)

Yakko


suppose that $ a,b,M\in{\mathbb{N}},a>1,b>1$,let $ S \equal{} \{(x,y)|x,y\in{\mathbb{N}},a^x \minus{} b^y \equal{} M\}$,prove or disprove that $ S$ is  finite.

You forget, that $ a>1,b>1$.
I think, that is open problem. It is easy followed from abc conjecture.

[quote="Rust"]You forget, that $ a > 1,b > 1$.
I think, that is open problem. It is easy followed from abc conjecture.[/quote]
oh yes!thanks Rust :)

The ABC Conjecture is not needed.  One can use Baker's Theorem on linear forms in logarithms.
See homework problem 2 at the end of the notes in
http://www.math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes8.pdf


but this has me stumped. What is the area of a circle inscribed in a 5,12,13 triangle?

Well you could always use the formula A=rs

A = area of triangle
r = radius of inscribed circle
s = semiperimeter of triangle

[quote="zserf"]What's a semiperimeter? Also if anyone has a method that doesn't use that it would be appreciated.[/quote]
semiperimieter=half the perimeter.

Your formula appears to be correct, but I haven't learned that and this problem is taken from a text book, so there must be a different way. Just for the record I'm not cheating on my homework or anything, I saw the problem and couldn't solve it which is rare for me so I thought I'd ask.

[hide] The radius of an inscribed circle in a right triangle [img]http://img.photobucket.com/albums/v475/footballrocks41237/right_triangle_image.png[/img] is $r=\frac{ab}{a+b+c}$.  So the radius has to be $\frac{5\times12}{5+12+13}$, or $\frac{60}{30}=2$, and the area is $2^{2}\pi$ or $4\pi$[/hide]

Or use the formula $r=\frac{a+b-c}{2}$, where a and b are the legs and c is the hypotenuse.

[quote="superwizard188"]um,

r = ab/(a+b+c)
r(a+b+c) = ab
r(a+b+c)/2 = ab/2

Look familiar? :P

Anyways, if you would like to derive the rs formula, draw tangents from the center O of the inscribed circle to the sides of the right triangle ABC. Let them be 'r'. Then draw OA, OB, and OC. You're left with 3 triangles with the heights already drawn for you. The area of each of these triangles is r*AB/2, r*BC/2, and r*AC/2. The area of the whole triangle is then r(AB + BC + AC)/2 which gives you the formula A = rs.[/quote]

Yeah I derived the $r=\frac{ab}{a+b+c}$ from $A=rs$...it's just easier for me to plug it in the equation...

Actually, the A=rs is very easy to prove. If you have drawn the triangle and the incircle, draw 3 segments, one from each of the triangle's vertices to the center of the incircle. Then, draw the 3 radii to the sides of the triangle.

Now, looking at the triangle, you have 3 smaller triangles, one with base a, one with base b, and one with base c, and all of them with height r, so you have A = (ar + br + cr) /2 or A = rs.


Find the sum of all real values of $ k$ such that the following system of equations has no solutions:
$ k^2x\minus{}2y\minus{}2z\equal{}4$
$ \minus{}2x\plus{}k^2y\minus{}2z\equal{}8$
$ \minus{}2x\minus{}2y\plus{}k^2z\equal{}12$

[quote="lingomaniac88"]Find the sum of all real values of $ k$ such that the following system of equations has no solutions:
$ k^2x \minus{} 2y \minus{} 2z \equal{} 4$
$ \minus{} 2x \plus{} k^2y \minus{} 2z \equal{} 8$
$ \minus{} 2x \minus{} 2y \plus{} k^2z \equal{} 12$[/quote]

[hide="Solution"]
Summing gives
\[ (k^2\minus{}4)(x\plus{}y\plus{}z) \equal{} (k\minus{}2)(k\plus{}2)(x\plus{}y\plus{}z) \equal{} 24\]
So $ k \neq \minus{}2, 2$. We now need to show all other values of $ k$ yield a set of solutions. Indeed, since $ 2x\plus{}2y\plus{}2z \equal{} \frac{48}{(k\minus{}2)(k\plus{}2)}$, adding to the first equation gives $ \Longrightarrow (k^2\plus{}2)x\equal{}4\plus{}\frac{48}{(k\minus{}2)(k\plus{}2)}$ and so forth for $ y$ and $ z$.

So $ \minus{}2 \plus{} 2 \equal{} \boxed{0}$.  :huh: [/hide]

That's right.  There's a one-liner solution, as well.  Can you find it?

Wow.. $ k^2$ appears, not $ k$, so obviously the solutions will add up to $ 0$. Nice  :lol:

Edit: For any positive solution for $ k$ then clearly the negative of that will also be a solution, and vice versa. So all possible values for $ k$ will add up to $ 0$.


analytic geometry
graphing lines
slope
calculus
derivative
algebra proposed
algebra

Let $(x,y)$ be the root of system: \begin{eqnarray*}x^{2}+y^{2}&>& 1, \\ x+y &\ge& x^{2}+y^{2}. \end{eqnarray*} Find the pair $(x,y)$ such that the sum $x+2y$ takes the greatest value.

There are three conditions:
1) Outside the circle $C_{1}$: $x^{2}+y^{2}=1$.
2) Inside or on the circle $C_{2}$: $x^{2}+y^{2}=x+y$, that is $(x-\frac{1}{2})^{2}+(y-\frac{1}{2})^{2}=\frac{1}{2}$.
3) Maximizing $k$ in $x+2y=k$, $y=\frac{-1}{2}x+\frac{k}{2}$.

There are two points on $C_{2}$ where the slope of the tangent line is $\frac{-1}{2}$, and they are the ones that optimize $k$ in $y=\frac{-1}{2}x+\frac{k}{2}$ inside or on the circle. One is the minimum and one is the maximum.
We'll see that, luckily, the maximum is outside $C_{1}$, and it will be the answer to our problem.

Using implicit differentiation with $C_{2}$, $(x-\frac{1}{2})dx+(y-\frac{1}{2})dy=0$.
We want $\frac{dy}{dx}=\frac{-1}{2}$ $\to$ $(y-\frac{1}{2})=2(x-\frac{1}{2})$.
Subst. in $C_{2}$, $5(x-\frac{1}{2})^{2}=\frac{1}{2}$.
The two solutions are $(x,y)=(\frac{1}{2}\pm\frac{1}{\sqrt{10}}, \frac{1}{2}\pm\frac{2}{\sqrt{10}})$.
The maximum $k=\frac{3}{2}+\frac{5}{\sqrt{10}}$ is at $\boxed{(x,y)=(\frac{1}{2}+\frac{1}{\sqrt{10}}, \frac{1}{2}+\frac{2}{\sqrt{10}})}$.
It's outside $C_{1}$ because $x^{2}+y^{2}=1+\frac{3}{\sqrt{10}}>1$.

Modified problem: :)

Being $0<r<2$, let $S$ be the set of points $(x,y)$ that are solutions to the system: \begin{eqnarray*}x^{2}+y^{2}&>& r, \\ x+y &\ge& x^{2}+y^{2}. \end{eqnarray*} Find the lowest upper bound of the sum $x+2y$ for every pair $(x,y)$ in $S$.

In the last post, we've seen the solution for every $r<1+\frac{3}{\sqrt{10}}$.
But, what about $1+\frac{3}{\sqrt{10}}\le r<2$?


What is remainder when the sum of all the possible numbers created with 1,2, and 3 is divided by 9?

Masoud Zargar

[hide]There are 6 possible ways to arrange. The sum of the digits is 6. 6 times that is 36, which is divisible by 9, and remainder is 0.[/hide]

Sorry, my problem was not mentioned correctly, so it did not make sense.  :blush: As a reult, your solution is incorrect.  Sorry. The problem is supposed to be: figure out the remainder when the sum of all the 3-digit numbers created using 1,2, and 3 (only once each number) is divided by 9.

Masoud Zargar

lol, your answer was correct for your first post, but your solution was incorrect. But for the next one, your solution was correct.  What a coincident.

okay, so my first solution was also a coincedence? this is pretty amazing.

I believe so.  Since I didn't mention any conditions for the number of digits in teh number, I don't know what you were solving for.  :P

Masoud Zargar


Let $ a,b,c$ be non-negative real numbers. Prove that

$ \frac{a^3\plus{}2abc}{a^3\plus{}(b\plus{}c)^3}\plus{}\frac{b^3\plus{}2abc}{b^3\plus{}(c\plus{}a)^3}\plus{}\frac{c^3\plus{}2abc}{c^3\plus{}(a\plus{}b)^3}\ge1$

Equality holds for $ a\equal{}b\equal{}c$ or $ abc\equal{}0$ :)

Hi, I'm not can_hang but because of being your younger brother, I think I can do it.
It won't make can_hang have to think because with the forms ineq above, expand kill it easily  :)
After expand, we can lost abc  :)

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers. Prove that

$ \frac {a^3 \plus{} 2abc}{a^3 \plus{} (b \plus{} c)^3} \plus{} \frac {b^3 \plus{} 2abc}{b^3 \plus{} (c \plus{} a)^3} \plus{} \frac {c^3 \plus{} 2abc}{c^3 \plus{} (a \plus{} b)^3}\ge1$

Equality holds for $ a \equal{} b \equal{} c$ or $ abc \equal{} 0$ :)[/quote]
We have
\[ 1 \minus{} \sum \frac {a^3}{a^3 \plus{} (b \plus{} c)^3} \equal{} \sum \left( \frac {a^3}{a^3 \plus{} b^3 \plus{} c^3} \minus{} \frac {a^3}{a^3 \plus{} (b \plus{} c)^3}\right) \equal{} \sum \frac {3a^3bc(b \plus{} c)}{(a^3 \plus{} b^3 \plus{} c^3)(a^3 \plus{} (b \plus{} c)^3)}
\]
Hence, it suffices to show that
\[ 2\sum \frac {a^3 \plus{} b^3 \plus{} c^3}{a^3 \plus{} (b \plus{} c)^3} \ge \sum \frac {3a^2(b \plus{} c)}{a^3 \plus{} (b \plus{} c)^3}
\]

\[ \Leftrightarrow \sum \frac {2a^3 \minus{} 3a^2(b \plus{} c) \plus{} 2b^3 \plus{} 2c^3}{a^3 \plus{} (b \plus{} c)^3} \ge 0
\]

\[ \Leftrightarrow \sum \frac {(a \minus{} b)(a^2 \minus{} 2ab \minus{} 2b^2) \plus{} (a \minus{} c)(a^2 \minus{} 2ac \minus{} 2c^2)}{a^3 \plus{} (b \plus{} c)^3} \ge 0
\]

\[ \Leftrightarrow \sum \frac {(a \minus{} b)^3 \minus{} (c \minus{} a)^3 \plus{} 3c^2(c \minus{} a) \minus{} 3b^2(a \minus{} b)}{a^3 \plus{} (b \plus{} c)^3} \ge 0
\]
It suffices to show that
\[ \sum \frac {c^2(c \minus{} a) \minus{} b^2(a \minus{} b)}{a^3 \plus{} (b \plus{} c)^3} \ge 0
\]
and
\[ \sum \frac {(a \minus{} b)^3 \minus{} (c \minus{} a)^3}{a^3 \plus{} (b \plus{} c)^3} \ge 0
\]
The first inequality is equivalent to
\[ \Leftrightarrow \sum (a \minus{} b)\left( \frac {a^2}{b^3 \plus{} (c \plus{} a)^3} \minus{} \frac {b^2}{a^3 \plus{} (b \plus{} c)^3}\right) \ge 0
\]
Finally, to finish the proof, we will show that if $ a \ge b$, then
\[ \frac {a^2}{b^3 \plus{} (c \plus{} a)^3} \ge \frac {b^2}{a^3 \plus{} (b \plus{} c)^3}
\]

\[ \Leftrightarrow a^5 \minus{} b^5 \ge b^2(c \plus{} a)^3 \minus{} a^2(b \plus{} c)^3
\]

\[ \Leftrightarrow a^5 \minus{} b^5 \ge a^2b^2(a \minus{} b) \plus{} c^3(b^2 \minus{} a^2) \plus{} 3c^2ab(b \minus{} a)
\]
which is obviously true since $ a \ge b$ and $ c\ge 0$.
And the second inequality is equivalent to
\[ \sum (a \minus{} b)^3\left( \frac {1}{a^3 \plus{} (b \plus{} c)^3} \minus{} \frac {1}{b^3 \plus{} (c \plus{} a)^3}\right) \ge 0
\]

\[ \Leftrightarrow \sum \frac {(a \minus{} b)^4(3c(a \plus{} b) \plus{} 3c^2)}{(a^3 \plus{} (b \plus{} c)^3)(b^3 \plus{} (c \plus{} a)^3)} \ge 0
\]
which is obviously true.
Our proof is completed. :)

P/s: Lam, thanks for your nice present, my friend. :) Best wishes for you :)

Very well ! Thanh you, Can !

In addition, we still have two another results, are also nice :

2/  $ \frac {a^3 \plus{} abc}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {b^3 \plus{} abc}{c^3 \plus{} a^3 \plus{} abc} \plus{} \frac {c^3 \plus{} abc}{a^3 \plus{} b^3 \plus{} abc}\ge2$

3/  $ \frac {a^3 \plus{} abc \plus{} b^3}{c^3 \plus{} (a \plus{} b)^3} \plus{} \frac {b^3 \plus{} abc \plus{} c^3}{a^3 \plus{} (b \plus{} c)^3} \plus{} \frac {c^3 \plus{} abc \plus{} a^3}{b^3 \plus{} (c \plus{} a)^3} \ge1$

[hide="@Can"] Firstly, congratulation you published successfully your book, my friend. Wish you'd be lucky, happy and always get success.
   Last period of time I who meet with some of difficulties, have whenever me to deem me will die all, but think about your admonitions, I have power and addition trust to continue to pace. You and I have quite homologous situations, ought to can sympathetically give together.I am always respectfully that. If I have thingummy cause the disappointment of you, I am apologetical. Once again, thankfully you is mickle![/hide]

3/ $ \sqrt {\frac {a^3}{b^3 \plus{} c^3 \plus{} abc}} \plus{} \sqrt {\frac {b^3}{c^3 \plus{} a^3 \plus{} abc}} \plus{} \sqrt {\frac {c^3}{a^3 \plus{} b^3 \plus{} abc}}\ge\sqrt3$ for all $ a,b,c>0$

It's stronger then a Nguyen Van Thach's following inequality
 $ \frac {a^3}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {b^3}{c^3 \plus{} a^3 \plus{} abc} \plus{} \frac {c^3}{a^3 \plus{} b^3 \plus{} abc}\ge1$

The following ineq is true, too.
$ \sum\ \sqrt[]{\frac{a^3}{4b^3\plus{}4c^3\plus{}abc}} \geq\ 1$

[quote="nguoivn"]The following ineq is true, too.
$ \sum\ \sqrt []{\frac {a^3}{4b^3 \plus{} 4c^3 \plus{} abc}} \geq\ 1$[/quote]
Try $ a\equal{}b\equal{}1,c\equal{}0.01$ ;)

[quote="can_hang2007"][quote="nguoivn"]The following ineq is true, too.
$ \sum\ \sqrt []{\frac {a^3}{4b^3 \plus{} 4c^3 \plus{} abc}} \geq\ 1$[/quote]
Try $ a \equal{} b \equal{} 1,c \equal{} 0.01$ ;)[/quote]

Equality occurs when $ a \equal{} b \equal{} c$ and $ a \equal{} b;c \equal{} 0$  :maybe:
So, at least, it's trues with $ c\equal{}0$

[quote="nguoivn"][quote="can_hang2007"][quote="nguoivn"]The following ineq is true, too.
$ \sum\ \sqrt []{\frac {a^3}{4b^3 \plus{} 4c^3 \plus{} abc}} \geq\ 1$[/quote]
Try $ a \equal{} b \equal{} 1,c \equal{} 0.01$ ;)[/quote]

Equality occurs when $ a \equal{} b \equal{} c$ and $ a \equal{} b;c \equal{} 0$  :maybe:
So, at least, it's trues with $ c \equal{} 0$[/quote]

Your problem's wrong for $ c \equal{} 0$, nguoivn  :wink:

With $ c\equal{}0$, it's becomes:
$ \frac{1}{2} (\sqrt[]{\frac{a^3}{b^3}}\plus{}\sqrt[]{\frac{b^3}{a^3}}) \geq\ 1$ (always trues)  :maybe:

The original inequality is a nice particular case of the inequality

$ \frac {a^3 \plus{} (p\plus{}2)abc}{a^3 \plus{} (b \plus{} c)^3\plus{}3pabc} \plus{} \frac {b^3 \plus{}(p\plus{}2)abc}{b^3 \plus{} (c \plus{} a)^3\plus{}3pabc} \plus{} \frac {c^3 \plus{}(p\plus{}2)abc}{c^3 \plus{} (a \plus{} b)^3\plus{}3pabc}\ge1$

which holds for $ \minus{}2\sqrt [3] 2 \le p\le 2$. The case $ p\equal{}\minus{}2$ is nice, and the case $ p\equal{}2$ is very nice.

[quote="nguoivn"]With $ c \equal{} 0$, it's becomes:
$ \frac {1}{2} (\sqrt []{\frac {a^3}{b^3}} \plus{} \sqrt []{\frac {b^3}{a^3}}) \geq\ 1$ (always trues)  :maybe:[/quote]

Yes, it's true for $ c\equal{}0$. But generally, it's wrong for $ c\sim 0$, example $ c\equal{}0.1$.

[quote="dduclam"]3/ $ \sqrt {\frac {a^3}{b^3 \plus{} c^3 \plus{} abc}} \plus{} \sqrt {\frac {b^3}{c^3 \plus{} a^3 \plus{} abc}} \plus{} \sqrt {\frac {c^3}{a^3 \plus{} b^3 \plus{} abc}}\ge\sqrt3$ for all $ a,b,c > 0$

It's stronger then a Nguyen Van Thach's following inequality
 $ \frac {a^3}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {b^3}{c^3 \plus{} a^3 \plus{} abc} \plus{} \frac {c^3}{a^3 \plus{} b^3 \plus{} abc}\ge1$[/quote]

I am sorry, the Nguyen Van Thach's inequality is $ \sum\frac {a^3}{a^3 \plus{} b^3 \plus{} abc}\ge1$
But above my result is still true  :)

And I hope the following general inequalities is also true:
1/ $ \frac {a^3}{b^3 \plus{} c^3 \plus{} kabc} \plus{} \frac {b^3}{c^3 \plus{} a^3 \plus{} kabc} \plus{} \frac {c^3}{a^3 \plus{} b^3 \plus{} kabc}\ge1; \forall a,b,c,k > 0$

2/ $ \sqrt {\frac {a^3}{b^3 \plus{} c^3 \plus{} kabc}} \plus{} \sqrt {\frac {b^3}{c^3 \plus{} a^3 \plus{} k abc}} \plus{} \sqrt {\frac {c^3}{a^3 \plus{} b^3 \plus{} kabc}}\ge\frac3{\sqrt{2\plus{}k}}$
$ \forall a,b,c > 0; k > 1$

[quote="Vasc"]The original inequality is a nice particular case of the inequality

$ \frac {a^3 \plus{} (p \plus{} 2)abc}{a^3 \plus{} (b \plus{} c)^3 \plus{} 3pabc} \plus{} \frac {b^3 \plus{} (p \plus{} 2)abc}{b^3 \plus{} (c \plus{} a)^3 \plus{} 3pabc} \plus{} \frac {c^3 \plus{} (p \plus{} 2)abc}{c^3 \plus{} (a \plus{} b)^3 \plus{} 3pabc}\ge1$

which holds for $ \minus{} 2\sqrt [3] 2 \le p\le 2$. The case $ p \equal{} \minus{} 2$ is nice, and the case $ p \equal{} 2$ is very nice.[/quote]

Nice, Vasc  :) I have to use ABC theorem to show it is rightly  :lol: