Let $A = $$\begin{bmatrix}2&0&x&-1\\1&3&y&0\\0&-1&z&-2\\z&z&xy/11&x\end{bmatrix}$
If $det(A) = x(2z+y+1)-z(y+1-z)$ when $x,y,z$ are distinct reals Find $\frac{3}{2}(x+z)-3y$

This one should be in College Playground/Superior Algebra/ Proposed problem


abstract algebra
superior algebra
superior algebra unsolved

Prove that if |G| = p^3, with p prime, then G is abelian or |Z(G)| = p.

THANKS!!!

Notation:

|A| is the order of A
Z(G) is the center of G

Suppose that $ G$ is nonabelian. Since, $ G$ is a $ p$-group, $ Z(G)$ is nontrival. There are two cases: $ |Z(G)|\equal{}p$, or $ |Z(G)|\equal{}p^2$.
The later cannot take place, since it implies $ G/Z(G)$ has order $ p$, which is a cyclic group. Consequently $ G$ is abelian, a contradiction.


Support
modular arithmetic
number theory
relatively prime

Assume that chicken nuggets come in packs of 6, 9,and 20.
Using various numbers of packs, what is the largest number of nuggets that you cannot buy?

[hide="Hint"]Base this off of the Chicken McNugget theorem, which supports two relatively prime variables. This is an extension[/hide]

[hide="Answer"]$ 43$...but I don't know how to show that...[/hide]

I wouldn't call this an extension of the Chicken McNugget Theorem: it's vaguely related, but isn't a direct consequence of it, nor requires knowledge of it.

[hide="hint that shows 123456789 has correct answer"]
Consider the number of chicken nuggets you want modulo 3.
[/hide]

The Chicken McNugget Theorem gives the solution to the [url=http://mathworld.wolfram.com/CoinProblem.html]coin problem[/url] for $ n \equal{} 2$.  An explicit solution is known for $ n \equal{} 3$, but it's very different.

Hmm last year in math club I saw this, I forgot how to solve it though...It has to do with modular arithmetic.

I understand that $ n\equiv 1 \pmod{3}$ and $ n\equiv 1 \pmod{3}$, meaning that $ n\equiv 1 \pmod{6}$, but I can't get past that...

Does it have anything to do with the fact that 49 is the first number where $ n\equiv 1 \pmod{6}$ that doesn't work?

Can someone outline exactly how to do it? Because I still have no idea.

I don't see where you're getting $ n\equiv 1\pmod{3}$ (well, what's $ n$?)

The reason mod 3's so important is that 6 and 9 are divisible by 3, so if $ m$ is the number of chicken nuggets you want, $ m$ determines the number of 20-packs modulo 3. Considering how big 20 is compared to 6 and 9, trying to order a number of nuggets equal to something like 17 or 22 (or even 43) has problems.

If you want to show more than 43 chicken nuggets are doable, try to order as few 20-packs as possible.

Here's how you would set it up if you're using $ \pmod{3}$:

Assume that we buy as few 20-packs as possible.  If we have three 20-packs, we can replace them for ten 6-packs, so we can't have more than three 20-packs.  Now we have three cases based on the number of 20-packs we have.

Case 1: No 20-packs - we can only get $ 0\pmod{3}$ nuggets.
Case 2: One 20-pack - we can only get $ 2\pmod{3}$ nuggets.
Case 3: Two 20-packs - we can only get $ 1\pmod{3}$ nuggets.

For each case we can find the largest number of nuggets that we can't get (in that residue class $ \pmod{3}$, since we can't get anything in a different residue class).  Now this problem turns into the simpler problem where we only have two sizes to choose from.

$ \pmod{2}$ also works for this problem in exactly the same way, see if you can figure it out.  :)


You can change your password under profile.  You'd want to post something like this here: [url]http://www.artofproblemsolving.com/Forum/index.php?f=144[/url]

[quote="ashpotter"]sorry if this in the wrong place  :?  :(  :blush: . where do you post questions about general things like where do you change your password and stuff[/quote]It's a good idea to first have a glance (if not even read :D) the [url=http://www.mathlinks.ro/Forum/faq.php]FAQ[/url] :P and then ask questions ;)


projective geometry
ratio
geometry unsolved
geometry

how does pappus theorem imply desargues theorem?

We denote the points $ S\equiv AC\cap A'C'$ and $ X\equiv AA'\cap CC'$ and $ Z\equiv A'C'\cap XB.$

Because of the points $ X\equiv AA'\cap CC'$ and $ B\equiv P'A'\cap R'C'$ and $ Z\equiv AP'\cap CR'$ are collinear, based on the [b][size=100]Desarques theorem[/size][/b], we have that the triangles $ \bigtriangleup AP'A'$ and $ \bigtriangleup CR'C'$ are perspective, where $ P'\equiv A'B\cap AZ$ and $ R'\cap BC'\cap CZ.$

So, the line segments $ AC,\ P'R',\ A'C',$ are concurrent at one point and then, the points $ S,\ P',\ R',$ are collinear.

We consider the triangles $ \bigtriangleup SAA',\ \bigtriangleup R'C'C$ and because of the collinearity of the points $ X\equiv AA'\cap C'C$ and $ B\equiv SA\cap R'C'$ and $ Z\equiv SA'\cap R'C,$ applying again the [b][size=100]Desarques theorem[/size][/b], we conclude that these triangles are perspective and so, the line segments $ SR',\ AC',\ A'C,$ are concurrent at one point. Hence, the point $ Q\equiv AC'\cap A'C,$ lies on the line segment $ P'R'.$

$ \bullet$ We consider now, the points $ A',\ B',\ Z,\ C',$ as the points of intersection of the pencils $ A.A'B'ZC',\ C.A'B'ZC'$ from the line segment $ SC'$ and aqpplying the [b][size=100]Double ratio[/size][/b] ( = [b][size=100]Cross Ratio[/size][/b] ) theory, we have that  $ (A.A'B'ZC') \equal{} (C.A'B'ZC')$ $ ,(1)$

But, $ (A.A'B'ZC') \equal{} (A',P,P',K)$ $ ,(2)$ and $ (C.A'B'ZC') \equal{} (L,R,R',C')$ $ ,(3)$ where $ K\equiv A'B\cap AC'$ and $ L\equiv BC'\cap A'C.$

From $ (1),\ (2),\ (3)$ $ \Longrightarrow$ $ (A',P,P',K) \equal{} (L,R,R',C')$ $ ,(4)$

From $ (4)$ we conclude that the points $ P,\ Q,\ R,$ as the [b][size=100]Pappus theorem[/size][/b] states, are collinear and the proof is completed.

Kostas Vittas.

what is double ratio theory? thanks vittasko

[quote="jaiawaj"]what is double ratio theory?[/quote]
All about the [b][size=100]Cross Ratio[/size][/b] theory ( see by google searching ).

In [b][size=100]Hellas[/size][/b] ( Greece ), [b][size=100]Double Ratio[/size][/b] ( = [b][size=100]Διπλός Λόγος[/size][/b] ), is more ordinary than [b][size=100]Cross Ratio[/size][/b].

Kostas Vittas.


AMC
AIME
USA(J)MO
USAJMO
USAMO
linear algebra
matrix

I have a few questions about the path to the IMO.

1.  What is the average score on AMC 10 and AIME and/or AMC 12 and AIME required to make USAJMO and/or USAMO.  (I know there is no statistics for USAJMO yet.)

2.  What happens after USAJMO?  Is there anything else?  Can you make it to an MOP from this?  Can you make it to USAMO from this?  Can you make it to IMO from this?

3.  What is MOP, and what are those colors that everyone talks about?  Red MOP, Black MOP, etc etc?

4.  What is the index for calculating USAMO qualification?

5.  How difficult is it to make USAMO/USAJMO?  How difficult is it to make AIME?

Sorry for all the questions, but I'm a bit confused.

1. See "Score Matrix for AIME/AMC1012" and "Index for AIME/AMC1012" statistics for 2007-2009 from [url=http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml]here[/url] to get a feel for it.

2. Problems 3 and 6 on the USAJMO is the same as problems 1 and 4 on the USAMO. Based on your score on those to problems, you can make Red MOP (only for 9th graders, about 25-30 people make it). So from USAJMO, you could get to Red MOP using problems 3/6 if you are in 9th grade. I don't think you can go anywhere else. The USAJMO was intended to be for experience rather than prestige.

3. MO(S)P stands for Mathematical Olympiad (Summer) Program. You can find a pretty good explanation [url=http://www.artofproblemsolving.com/Wiki/index.php/Mathematical_Olympiad_Summer_Program]here[/url]. Basically, it is a 3-week summer camp that you qualify to through AMC/AIME/USA(J)MO. Black MOP is the highest level, and the top 12 people from the USAMO qualify. From these 12 people, the US IMO team is selected. Blue MOP is the next level, with 12-18 non-seniors qualifying. For Red MOP, see #2. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294573]This[/url] thread has more info.

4. AMC 12 score + 10*AIME score = USAMO index. The USAJMO index is calculated the same way, except it's your AMC 10 score rather than AMC 12.

5. About 270 people make USAMO and 230 make USAJMO (this will be based on others score, so there is no permanent cutoff). Last year, about 10,000 people made it to AIME, but it was lower previous years (and there is a permanent cutoff for this).

Alright, thanks.  :)

That was some helpful info.


[quote="GoKorea"]In triangle ABC satisfying (a^4+b^4+c^4)/(sin^4A+sin^4B+sin^4C)=81

find maximum value of "a"

help;;[/quote]

from rule of sin

a/sinA=b/sinB=c/sinC=2r

(a^4+b^4+c^4)/(sin^4A+sin^4B+sin^4C)=16r^4

from the  relationship given

81=16r^4===== r=3/2

a/sinA=2r

when A=90

a= 3


Prove that

(n^2000 - 1 ) / (n-1)   cannot be a perfect square for all natural number [b]n>1[/b]

[quote="ehsan2004"]When $n\geq 1,\frac{n^{2000}-1}{n-1}=(n^{1000}+1)(n^{500}+1)(\frac{n^{500}-1}{n-1})$.We show that $A= n^{1000}+1 , B= n^{500}+1, C=\frac{n^{500}-1}{n-1}$
Then $B\mid A-2 , C\mid A-2$ and $C\mid B-2$. Then $gcd(A,B)=gcd(B,C)=gcd(C,A)\leq 2$so that the products of numbers $A,B,C$ be the perfect square, then any of them be the perfect square or it's should be the double of a perfect square. It' obviously that $A,B$ are not perfect squares . Suppose that any of the numbers $A,B$ be the double of the perfect square so that $4AB$ is the perfect square too but $4AB=4n^{1500}+4n^{1000}+4n^{500}+4$ then we know that $n\geq 1$ and $(2n^{750}+n^{750})^2 < 4AB < (2n^{750}+n^{750}+1)^2$ so that $4AB$ is not the perfect square. ;)[/quote]

Can you explain more detail why gcd(A,B)=gcd(A,C)=gcd(B,C) . After you proof A , B , 4AB cannot be prefect square , I still cant see why ABC is not perfect square  :oops:


[tex]a,b,c>0[/tex].Prove that :
[tex]\large \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}} \le \frac{5\sqrt{a+b+c}}{4}[/tex].

It should be:

$\sum\frac{a}{\sqrt{a+b}} \le \frac{5\sqrt{a+b+c}}{4}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=1459 .

  darij


algebra
polynomial
modular arithmetic
probability
number theory
number theory open

Let $ P(x,y)$ be a polynomial with integer coefficients, in 2 variables. Is there a general approach to find all natural n's for which the equation $ P(x,y) \equiv 0\pmod n$ has no solutions in $ Z/nZ$?

The motivation - it is known that the equation $ y^5 \minus{} x^2 \minus{} 4 \equiv 0 \pmod {11}$ has no solutions ([url]http://www.mathlinks.ro/viewtopic.php?t=150625[/url]). I wrote a simple program that checks whether positive integers, other than 11 (and less than ~500), also have the same property, and it turns out that the answer is no. I speculate that 11 is the only one, but have no idea how to approach it.

There is a very specific approach that I can describe.

[b]Prove or disprove:[/b]  Let $ p$ be a prime and let $ p \minus{} 1 \equal{} ab$ where $ a, b > 1$.  Then there exists $ k$ such that $ x^a \plus{} y^b \equiv k \bmod p$ has no solutions.

For example, this is true of $ x^3 \plus{} y^4 \equiv 19^{19} \bmod 13$.

These types of Diophantine equations are special because it is very easy to count the number of distinct possible values $ x^a$ can take $ \bmod p$.  If you are not aware of this lemma,

[b]Prove:[/b]  When $ p$ does not divide $ x$, the quantity $ x^a \bmod p$ can take on $ \frac{p\minus{}1}{\gcd(a, p\minus{}1)}$ distinct values.

[b]an idea[/b]
it's just had an idea. although nice one. i would like to hear opinion about it because i don't really know anything about probabilistic number theory and it just came into my mind. it is not really a proof of anything but is a method that can help us choose our moduls.
i'll show the method on this example $ x^5 \equal{} y^2 \plus{} 4$

obviously we're interested in prime moduls. we want to find such prime moduls that would give no solutions. first because $ x^5 \mod p$ can take precisely $ \frac {p \minus{} 1}{gcd(5,p \minus{} 1)} \plus{} 1$ distinct values (i counted 0 as well) we can see that if $ p$ is not of a form $ 5k \plus{} 1$ than $ x^5$ can take any value $ \mod p$ meaning there is no way it would give no solutions for the equation. therefore we further focus ourselves to primes of a form $ p \equal{} 10k \plus{} 1$ (5k+1 and obviously odd)(in this particular case if we just want to find a modul it's obvious to first try $ p \equal{} 11$ at this point but see further atempt to try to find more such moduls)

there are now $ 2k \plus{} 1$ diferent values that $ x^5$ can reach and $ 5k \plus{} 1$ such that $ y^2$ and surely $ y^2 \plus{} 4$ can $ \mod p$.
the probabillity of two subsets(of powers $ 2k \plus{} 1$ and $ 5k \plus{} 1$) of the set of reminders modulo $ 10k \plus{} 1$ not to intersect each other at any point is $ \frac {\binom{10k \plus{} 1}{2k \plus{} 1}\binom{8k}{5k \plus{} 1}}{\binom{10k \plus{} 1}{2k \plus{} 1}\binom{10k \plus{} 1}{5k \plus{} 1}}$ (the number of posibilities where they don't intersect devided by all posibilities) which turns out to be equal to
$ \frac {(8k)!\cdot (5k)!}{(3k \minus{} 1)!\cdot (10k \plus{} 1)!}$

so it turns out that this "probability" of $ 11$ beeing such a modul turns out to be $ \frac {2}{33}$ or $ 0.0606061$
of $ 31$ (which is the next prime number of desired form) $ \frac {11}{4495}$ or $ 0.00244716$
while for example of $ 101$ is $ \frac {1480}{56666053737}$ or $ 0.0000000261179$
so if u checked all moduls up to $ 500$ and found none i think it's higly unlikely any other exist

You may want to look up Hasse's theorem. In principle it says: a general polynomial of at least $ 2$ variables archieves some given value $ \mod p$ for all but finitely many $ p$.
Additionally, Hensel's lemma garantees you almost always solutions $ \mod p^n$ if there is one $ \mod p$.

Thus you are likely to expect solutions.


trigonometry
calculus
calculus computations

Solve $ \left\{
\begin{array}{ll}
\frac{dy}{dt}\equal{}y\plus{}z\minus{}\sin t &\quad  \\
\frac{dz}{dt}\equal{}\minus{}y\plus{}z\minus{}\cos t &\quad
\end{array}
\right.$

hello, i have found
$ y(t)\equal{}C_1e^t\cos(t)\plus{}C_2e^t\sin(t)\plus{}\sin(t)$
$ z(t)\equal{}\minus{}C_1e^t\sin(t)\plus{}C_2e^t\cos(t)\plus{}\cos(t)$.
Sonnhard.


$f'(z)=\cos z$
$f''(z)=-\sin z$
$f'''(z)=-\cos z$
$f''''(z)=\sin z$

You can repeat this process; the result should be clear.


:roll: What is the product 1*2*3*4*.........n?

This does not really belong to this forum..
And $1 \cdot 2 \cdot \ldots \cdot n = n!$

EDIT : Nevermind : http://www.mathlinks.ro/Forum/viewtopic.php?t=74971


[hide="Watch the answer for THE ULTIMATE QUESTION to Life, the Universe, and Everything."]
[youtube]5GUV7zz-8Oc[/youtube]
Never expected that did'ya[/hide]

^^ LOL ^^

And actually I did expect that. :P

[hide=so did I. I read the book.][/hide]

[hide="Hmmm"][b]@ above poster:[/b] No cheating!  :ninja: [/hide]


inequalities
function
symmetry
calculus
derivative
algebra
rational function

[b]Problem 1[/b] [ Đào Hải Long] Let $ a,b,c$ be real numbers. Find the best $ k$ constant such that the following inequality is correct:
\[ \frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{(b\minus{}c)^2}\plus{}\frac{1}{(c\minus{}a)^2}\ge \frac{k}{a^2\plus{}b^2\plus{}c^2}\]

Above inequality was appeared in Mathlinks Forum long time ago (maybe it was posted by hungkhtn)
And we have a stronger one with $ a,b,c\ge 0$

[b]Problem 2[/b]  Let $ a,b,c$ be positive real numbers.
Find the best $ k$ constant such that the following inequality is correct:
\[ \frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{(b\minus{}c)^2}\plus{}\frac{1}{(c\minus{}a)^2}\ge \frac{k}{ab\plus{}bc\plus{}ca}\]

We can solve both of them by same method. Let try your best  :)

Let us say $ a > b > c$
fix $ a \minus{} b \equal{} x,b \minus{} c \equal{} y$, and consider the function $ LHS \minus{} RHS \equal{} f(x,y,c)$
then since $ a^2 \plus{} b^2 \plus{} c^2,ab \plus{} bc \plus{} ca$ are both increasing function on $ c$
enough to consider $ c \equal{} 0$
from then on by homogenousity and symmetry enough to consider $ b \equal{} 1$
from then on the problem reduces to finding the extreme values of a rational function
easy by derivative

k=4. It is problem 6.27 in Pham Van Thuan's inequality book.

[quote="zaizai-hoang"][b]Problem 1[/b] [ Đào Hải Long] Let $ a,b,c$ be real numbers. Find the best $ k$ constant such that the following inequality is correct:
\[ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2}\ge \frac {k}{a^2 \plus{} b^2 \plus{} c^2}
\]
Above inequality was appeared in Mathlinks Forum long time ago (maybe it was posted by hungkhtn)
And we have a stronger one with $ a,b,c\ge 0$

[b]Problem 2[/b]  Let $ a,b,c$ be positive real numbers.
Find the best $ k$ constant such that the following inequality is correct:
\[ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2}\ge \frac {k}{ab \plus{} bc \plus{} ca}
\]
We can solve both of them by same method. Let try your best  :)[/quote]

I think the useful method to prove 2 inequalities above is using Mixing Variable in the form: f(a-c,b-c,0) with c=min(a,b,c)>=0 since at that time, LHS is still the same but the RHS is increasing (since a^2+b^2+c^2>(a-c)^2+(b-c)^2).Then proving when c=0 and consider the case when a=b to guess the best k constant or using this substitution: a^2+b^2=((a-b)^2+(a+b)^2)/2)


[quote="chien than"]Prove that:
$(\frac{3}{x}-1)(\frac{3}{y}-1) (\frac{3}{z}-1) \geq 512$
where $x;y;z$ are positivereal numbers satisfying $x+y+z=1$[/quote]
$\left[{\left({\frac{3}{x}-1}\right)\left({\frac{3}{y}-1}\right)\left({\frac{3}{z}-1}\right)}\right]^{2}\geq\frac{{64}}{{27}}\left[{\left({\frac{3}{{x+y}}-1}\right)\left({\frac{3}{{y+z}}-1}\right)\left({\frac{3}{{z+x}}-1}\right)}\right]^{2}$


[quote]The power source will be a 6 V lantern battery (preferably one that has not been stored for a long time or used much) and the circuit will also contain an ammeter to measure the current flow. The current can be controlled with a variable resistor, or by adjusting the separation of the electrodes. The students should select a current of no more than 2.5 A.[/quote]

This is an except from the description of one of the events of a local CHEMATHON competition. Our task is to engineer a method of providing a constant current over time. Are there any suggestions for such methods?

Or alternatively, how do the separation of electrodes or use of a variable resistor control a constant flow?

u can always reduce the resistance thru a rheostat and keep the current constant :P


limit
trigonometry
integration
calculus
calculus computations

Compute the limit:

$\displaystyle \lim_{n \to \infty} \arctan \frac{1}{n} \cdot \sum_{k=1}^{n} \frac{1}{1+ \tan \frac{k}{n}}.$

Riemann sum $\int_{0}^{1}\frac{1}{1+\tan x}dx$


Given a set of 51 integers between 1 and 100 (inclusive), show that at least one member of the set must divide another member of the set.

I think it's in Paul Zeitz's book, where 100 is replaced by 2n and 51 is replaced by n+1.

A typical pigeonhole principle problem.

there is also a nice induction proof, find it!


calculus integration trigonometry limit Putnam logarithms function

[quote="skylarth"]1) $\int_0^ \infty \frac{\tan^{-1}{\pi x}-\tan^{-1}{x}}{x} dx$

2) Show that $\int_0^ 1 \int_0^ 1 \int_0^ 1 \frac{dxdydz}{1-xyz} = \sum_{n=1}^\infty \frac{1}{n^3}$[/quote]
The first one I believe can be done by dividing both terms in the numerator by x then you have to use the part integration method.  I may be wrong though; I don't have the time to do it out now.

The second one can be done a couple of ways.  Here is the way I like:

$\sum_{n=1}^{\infty}\frac{1}{n^{3}}=\sum_{n=0}^{\infty}\frac{1}{\left(n+1\right)^{3}}$

$\int_{0}^{1}x^{k}\,dx=\frac{1}{k+1}\implies \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left(xyz\right)^{k}\,dx\,dy\,dz=\frac{1}{\left(k+1\right)^{3}}$

$\sum_{n=0}^{\infty}\frac{1}{\left(n+1\right)^{3}}=\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left(xyz\right)^{k}\,dx\,dy\,dz$

Then switching the sum and the integrals, you get:

$\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left(xyz\right)^{k}\,dx\,dy\,dz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}\left(xyz\right)^{k}\,dx\,dy\,dz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xyz}\,dx\,dy\,dz$

I don't believe so.  In fact, I think only even powers have been computed (such as zeta[2] and zeta[4]).

[quote="perfect_radio"]is there any easy way to compute $\displaystyle \sum \frac{1}{n^3}$?[/quote]

Well, I don't know anyone who has computed that. I know only that it has been proved to be an irrational number. (I assume you mean $n$ from 1 to $\infty$.)

Just that no one has done it yet doesn't mean that it is impossible   ;)

I'm not sure about perfect radio's question at the moment.

I couldn't solve the first problem using integration by parts, so here is an alternative.

$\int_0^ \infty \frac{\tan^{-1}{\pi x}-\tan^{-1}{x}}{x} dx = \int_0^ \infty \frac{\tan^{-1}{\pi x}}{x} dx - \int_0^ \infty \frac{\tan^{-1}{x}}{x} dx = F(\pi) - F(1)$

,where $F(k) = \int_0^ \infty \frac{\tan^{-1}{k x}}{x} dx$

$\frac{d}{dx} \tan^{-1} {kx} = \frac{k}{1+k^2 x^2} \rightarrow \tan^{-1} {kx} = \int_0^ \infty \frac{d}{dx} \tan^{-1} {kx} dx = \int_0^ \infty \frac{k}{1+k^2 x^2} dx = \int_0^ \infty \sum_{n=0}^ \infty (-1)^n k^{2n+1} x^{2n} dx = \sum_{n=0}^ \infty \frac{(-1)^n k^{2n+1} x^{2n+1}}{2n+1} \vert_{0}^ \infty$
$= \lim_{i\to \infty} \sum_{n=0}^{\infty} \frac{(-1)^n k^{2n+1} x^{2n+1}}{2n+1} = G(k)$

$F(k) = \int_0^ \infty G(k) dx$

This just became a big mess, I did something wrong.

I will do the second integral, as the same method can be applied to the first:

$\int_{0}^{\infty}\frac{\arctan{x}}{x}\,dx=\int_{0}^{\frac{\pi}{2}}\tan{u}\,du+\int_{0}^{\frac{\pi}{2}}\cot{u}\,du=\infty$

This was obtained by making the substitution u=arctan(x).  But this integral diverges.  Are you sure you copied the problem correctly?

Edit:  When I calculated your first problem on my computer, I came up with 1.79813749986.

[quote="perfect_radio"]okay, so it's not possible. but how to prove that $\sum \frac{1}{n^3} = \frac{5}{2} \sum \frac{(-1)^{k-1}}{k^3 \binom{2k}{k}}$, where sums are from $1$ to $\infty$.[/quote]
This formula was used by apery to prove that $\zeta(3)$ is irrational. http://mathworld.wolfram.com/AperysConstant.html

For 1), you can proceed as with this year's Putnam A5

Put $I(a)=\int^\infty_0 \frac{\arctan at-\arctan t}{t}dt$, $a\geq 1$.

Then

$\begin{eqnarray*} I'(a)&=&\int^\infty_0 \frac{\partial}{\partial a} \left(\frac{\arctan at-\arctan t}{t}\right)dt \\ &=&\int^{\infty}_0 \frac{1}{a^2t^2+1}dt \\ &=&\frac{\pi}{2a}$

It follows that $I(a)=\frac{\pi}{2}\log a+C$

Since $I(1)=0$, we conclude that $C=0$

So the answer is $I(\pi)=\frac{\pi}{2}\log \pi$

blahblahblah:

Your steps make sense, but $\frac{\pi}{2}\log{\pi}$ is different than the answer my computer gives when it is calculated numerically.  I don't think my computer is wrong, but I could have entered it wrong.  Can anyone else confirm?

According to Maple 10 (Classic Worksheet version, thankyouverymuch)

> int((arctan(Pi*x)-arctan(x))/x,x=0...infinity);

                            1/2 Pi ln(Pi)

>

When blahblahblah wrote $\log {\pi}$ he meant $\ln {\pi}$.

Kalle, partial differentiation is a method in multivariable calculus.

I always write the natural logarithm as $\log$.  I have 7 analysis textbooks within my arm's reach (Lax, Folland, Conway, Ahlfors, and Rudin, Rudin, and Rudin) and they all use $\log$ to denote the natural logarithm - that's good enough for me.

Let me rephrase that.

Sometimes $\log x$ is used to refer to $\log_{10} x$ and sometimes it used to refer to $\log_e x$.

$\ln x$ refers to $log_e x$ only.

The reason people might be getting different valued solutions using a numerical solution is because $\log {\pi}$ is used to mean $\log_e {\pi} \Longleftrightarrow \ln {\pi}$

Applause for using one of the solutions to A5, my favorite Putnam problem (even though I'm not in college yet).

Yes, well, I'm just glad that I got A5 this year  :D

[quote="Kalle"]I don't understand those solutions at all, blahblahblah. How can you have the integral as a function of a? Does this have something to do with multivariate calculus?[/quote]

He did it just like i did in calculus problem marathon. ( thanks to FMako has explained that :)
I also solve this problem using exactly the same method with blahblahblah's. But as usual you guys always online in midnight (indonesia Time) so i always late to post my solution. :lol:

[quote="perfect_radio"]how to prove that $\sum \frac{1}{n^3} = \frac{5}{2} \sum \frac{(-1)^{k-1}}{k^3 \binom{2k}{k}}$, where sums are from $1$ to $\infty$.[/quote]
Let me try: $\sum_{k=1}^K\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots (x+a_k)}=\frac{1}{x}-\frac{a_1a_2\cdots a_{K-1}}{(x+a_1)\cdots (x+a_K)}$ so the summand is of the form $A_0-A_k$ where $A_k=\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots(x+a_k)}$. Each term in the left hand side is equal to $A_0-A_k$. Now substituting $x=n^2$ and $a_k=-k^2$ and taking $k\leq K\leq n-1$ we have
\begin{eqnarray*} \sum_{k=1}^{n-1}\frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\cdots (n^2-k^2)}&=&\frac{1}{n^2}-\frac{(-1)^{n-1}(n-1)!^2}{n^2(n^2-1^2)\cdots (n^2-(n-1)^2)}\\ &=&\frac{1}{n^2}-\frac{2(-1)^{n-1}}{n^2\binom{2n}{n}}\end{eqnarray*}
Writing $C_{n,k}=\frac{1}{2}\frac{k!^2(n-k)!}{k^3(n+k)!}$ we get
$(-1)^kn(C_{n,k}-C_{n-1,k})=\frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\cdots (n^2-k^2)}$
from which we get the identity
\begin{eqnarray*} \sum_{n=1}^{N}\sum_{k=1}^{n-1}(-1)^k(C_{n,k}-C_{n-1,k})&=&\sum_{n=1}^{N} \frac{1}{n^3}-2\sum_{n=1}^N \frac{(-1)^{n-1}}{n^3\binom{2n}{n}}\\ &=&\sum_{k=1}^N(-1)^k(C_{N,k}-C_{k,k})\\ &=&\sum_{k=1}^{N}\frac{(-1)^k}{2k^3\binom{N+k}{k}\binom{N}{k}}+\sum_{k=1}^N\frac{(-1)^{k-1}}{k^3\binom{2k}{k}}\\ \end{eqnarray*}
Now the identity follows since
$\sum_{k=1}^{N}\frac{(-1)^k}{2k^3\binom{N+k}{k}\binom{N}{k}}\to 0$
as $N\to\infty$.

(How can I use eqnarray-environment in AoPS?)

blahblahblah is correct for 1. $\frac{\pi}{2}\ln \pi$

another way to do 1 is:

$\int_0^ \infty \frac{\tan^{-1}{\pi x}-\tan^{-1}{x}}{x} dx = \int_0^ \infty \int_1^{\pi} \frac{1}{1+(xt)^2} dtdx = \int_1^{\pi} \int_0^ \infty \frac{1}{1+(xt)^2} dxdt= \int_1^{\pi} \frac{\tan^{-1} \infty - \tan^{-1} 0}{t} dt = \frac{\pi}{2} \int_1^{\pi} \frac{1}{t} dt = \frac{\pi}{2} \ln \pi$

hope i didnt make any mistakes.

It's correct.  My two favorite methods for integrating a difficult integral are introducing an extra parameter (as I did there) and changing a single integral into a double integral and using Fubini's theorem (as you did).  I find it very interesting that they lead to the same steps a lot of the time.


Let $ x,y,z$ are positive numbers such that $ x \plus{} y \plus{} z \equal{} 3$ then prove that
$ \sum {\sqrt {x \plus{} 3} \ge 6}$
By setting $ a \equal{} \sqrt {x \plus{} 3} ,b \equal{} \sqrt {y \plus{} 3} ,c \equal{} \sqrt {z \plus{} 3}$
We will prove the follow inequality
Let $ a,b,c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 6$ then prove that
$ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)} \le 3$
$ \Leftrightarrow q^2 \plus{} 12q \minus{} 12r \minus{} 192 \le 0$
By Schur ineq we have $ r \ge \frac {{8q \minus{} 72}}{3}$.
If $ q \ge 9$
$ \Leftrightarrow q^2 \plus{} 12q \minus{} 12r \minus{} 192 \le 0$
$ q^2 \plus{} 12q \minus{} 12r \minus{} 192 \le q^2 \plus{} 12q \minus{} 4\left( {8q \minus{} 72} \right) \minus{} 192$
$ q^2 \minus{} 20q \plus{} 96 \equal{} \left( {q \minus{} 12} \right)\left( {q \minus{} 8} \right) \le 0$
Which is true. In the case $ q \le 9$
$ q^2 \plus{} 12q \minus{} 12r \minus{} 192 \le \minus{} 12r \minus{} 3 < 0$
So we have done and it holds for $ a \equal{} b \equal{} c \equal{} 2$

[quote="anas"](a,b,c)>0 , such that ab+ac+bc=3
Prove that :

$ \sum {\sqrt {a \plus{} 3} } \geq 6$[/quote]
Let $ f(a,b,c) \equal{} \sqrt {a \plus{} 3} \plus{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \minus{} 6,$  $ g(a,b,c) \equal{} ab \plus{} ac \plus{} bc$ and $ a \equal{} \min\{a,b,c\}.$
Hence, $ g(a,b,c) \equal{} g\left(a,\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a,\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a\right)$ and
$ f(a,b,c) \minus{} f\left(a,\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a,\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a\right) \equal{}$
$ \equal{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \minus{} 2\sqrt {\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a \plus{} 3} \equal{}$
$ \equal{} \frac {b \plus{} c \plus{} 4a \minus{} 6 \plus{} 2\sqrt {(b \plus{} 3)(c \plus{} 3)} \minus{} 4\sqrt {(a \plus{} b)(a \plus{} c)}}{\sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \plus{} 2\sqrt {\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a \plus{} 3}} \equal{}$
$ \equal{} \frac { \minus{} b \minus{} 3 \minus{} c \minus{} 3 \plus{} 2\sqrt {(b \plus{} 3)(c \plus{} 3)} \plus{} 2b \plus{} 2c \plus{} 4a \minus{} 4\sqrt {(a \plus{} b)(a \plus{} c)}}{\sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \plus{} 2\sqrt {\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a \plus{} 3}} \equal{}$
$ \equal{} \frac {2\left(\sqrt {a \plus{} b} \minus{} \sqrt {a \plus{} c}\right)^2 \minus{} \left(\sqrt {b \plus{} 3} \minus{} \sqrt {c \plus{} 3}\right)^2}{\sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \plus{} 2\sqrt {\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a \plus{} 3}} \equal{}$
$ \equal{} \frac {(b \minus{} c)^2\left(\frac {2}{\left(\sqrt {a \plus{} b} \plus{} \sqrt {a \plus{} c}\right)^2} \minus{} \frac {1}{\left(\sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3}\right)^2}\right)}{\sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \plus{} 2\sqrt {\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a \plus{} 3}}\geq0.$
Thus, it's remains to check out that $ f(a,b,b)\geq0,$ where $ 2ab \plus{} b^2 \equal{} 3$ and $ 1\leq b\leq\sqrt3.$

$ \sum (a^2\minus{}3)(b^2\minus{}3) \leq 3 .$
 (   because  we  only test   for  some  value $ a,b,c$)
   use $ S.O.S$
$ 3\sum a^2b^2 \minus{} (\sum ab)^2 \leq 6 \sum a^2 \minus{}72\plus{}12 \sum a^2 \minus{} (\sum ab)^2$
  notice that : $ a\plus{}b\plus{}c\equal{}6$ hence $ ab\plus{}bc\plus{}ca  \leq 12$
$ \to  \sum (8\minus{}a^2)(b\minus{}c)^2 \geq 0$
  from  supposition   we have : $ a,b,c \geq \sqrt{3}$   therefor   $ a,b,c \leq 6\minus{} 2\sqrt{3}  < 2\sqrt{2}$

[quote="arqady"][quote="Honey_S"]Let $ x,y,z$ are positive numbers such that $ x \plus{} y \plus{} z \equal{} 3$ then prove that
$ \sum {\sqrt {x \plus{} 3} \ge 6}$
By setting $ a \equal{} \sqrt {x \plus{} 3} ,b \equal{} \sqrt {y \plus{} 3} ,c \equal{} \sqrt {z \plus{} 3}$
We will prove the follow inequality
Let $ a,b,c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 6$ then prove that
$ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)} \le 3$
[/quote]
Why it is enough? Thank you![/quote]

it calls 'phan chung' in Vietnamese but I do not know how it is in English. We change the position between the condition and the ineq.
In the ineq $ a \plus{} b \plus{} c \ge 6$ the condition is $ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)} \equal{} 3$
When we decrease for $ a \plus{} b \plus{} c \equal{} 6$ so $ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)}$ will decrease also and thus we will prove $ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)} \le 3$

[quote="Hong Quy"]$ \sum (a^2 \minus{} 3)(b^2 \minus{} 3) \leq 3 .$
 (   because  we  only test   for  some  value $ a,b,c$)
   [/quote]
I don't understand you. :(  Why
[quote="Hong Quy"]
$ \sum \sqrt {x \plus{} 3} \ge 6$   [/quote]
 is true? :wink:
[quote="Honey_S"]We change the position between the condition and the ineq.
In the ineq $ a \plus{} b \plus{} c \ge 6$ the condition is $ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)} \equal{} 3$
When we decrease for $ a \plus{} b \plus{} c \equal{} 6$ so $ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)}$ will decrease also and thus we will prove $ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)} \le 3$[/quote]
Why it is enough for the proof of our inequality?

The stronger inequality holds for $ ab \plus{} bc \plus{} ca \equal{} 3$:

$ (a \plus{} 3)(b \plus{} 3)(c \plus{} 3) \ge 64$.

These are sharper:

$ (2a \plus{} 3)(2b \plus{} 3)(2c \plus{} 3) \ge 125$;

$ \sqrt{5a\plus{}2}\plus{}\sqrt{5b\plus{}2}\plus{}\sqrt{5c\plus{}2} \ge 3\sqrt{7}$.

my idea about the solution  :blush: :

putting $ x^2\equal{}ab,y^2\equal{}ac,z^2\equal{}bc$

we get : $ a\equal{}\frac{{xy}}{z}, b\equal{}\frac{{xz}}{y} ,c\equal{}\frac{{yz}}{x}$
so the inéquality becomes:

$ \sum {\sqrt {\frac{{xy}}{z} \plus{} 3} }  \geq 6$

squaring we get:

$ \sum {\frac{{xy}}{z} \plus{} 9}  \plus{} 2\sum {\sqrt {\frac{{xy}}{z} \plus{} 3} } \sqrt {\frac{{xz}}{y} \plus{} 3}  \geq  36$

by cauchy :

$ \sum {\sqrt {\frac{{xy}}{z} \plus{} 3} } \sqrt {\frac{{xz}}{y} \plus{} 3}  \ge \sum {(x \plus{} 3) \equal{} \sum x }  \plus{} 9$

$ \sum {\frac{{xy}}{z}}  \plus{} 2\sum {x \geq 9}$

$ \Leftrightarrow \left( {\sum {xy} } \right)^2  \geq  9xyz$  but I'm blocked here  :blush:  :blush:
My second Idea :
the inequality  is equivalent to:

$ \sum {\frac{{(a^2  \minus{} 1)}}{{(a \plus{} 1)(\sqrt {a \plus{} 3}  \plus{} 2)}}}  \geq  0$
well we have to find an expression $ f(a,b,c)>\equal{}0$ so that:

$ \sum {\frac{{(a^2  \minus{} 1)}}{{(a \plus{} 1)(\sqrt {a \plus{} 3}  \plus{} 2)}}}  \ge \sum {\frac{{(a^2  \minus{} 1)}}{{f(a,b,c)}}}$
and :
$ \sum {\frac{{(a^2  \minus{} 1)}}{{f(a,b,c)}}}  \equal{} \frac{1}{{f(a,b,c)}}(\sum {a^2  \minus{} \sum {ab} } ) \geq 0$  :ninja:  :roll:

[quote="Vasc"]The stronger inequality holds for $ ab \plus{} bc \plus{} ca \equal{} 3$:

$ \sqrt {5a \plus{} 2} \plus{} \sqrt {5b \plus{} 2} \plus{} \sqrt {5c \plus{} 2} \ge 3\sqrt {7}$.[/quote]
The minimal $ k$ for which an inequality $ \sqrt {a \plus{} k} \plus{} \sqrt {b \plus{} k} \plus{} \sqrt {c \plus{} k} \ge 3\sqrt {1\plus{}k}$ holds for all non-negative $ a,$ $ b$ and $ c$ such that $ ab\plus{}ac\plus{}bc\equal{}3$ is

$ k_{min}\equal{}\max_{[1,\sqrt3]}\frac{9b^2\minus{}10b\plus{}1}{16b}\equal{}\frac{14\minus{}5\sqrt3}{8\sqrt3}\equal{}0.385...$

[quote="Honey_S"][quote="arqady"][quote="Honey_S"]Let $ x,y,z$ are positive numbers such that $ x \plus{} y \plus{} z \equal{} 3$ then prove that
$ \sum {\sqrt {x \plus{} 3} \ge 6}$
By setting $ a \equal{} \sqrt {x \plus{} 3} ,b \equal{} \sqrt {y \plus{} 3} ,c \equal{} \sqrt {z \plus{} 3}$
We will prove the follow inequality
Let $ a,b,c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 6$ then prove that
$ \sum {\left( {a^2 \minus{} 3} \right)\left( {b^2 \minus{} 3} \right)} \le 3$
[/quote]
Why it is enough? Thank you![/quote]

it calls 'phan chung' in Vietnamese but I do not know how it is in English. We change the position between the condition and the ineq.
 [/quote]

Arqady,this is [b]Contradiction[/b] method. If you can't understand the principal of this method,have a look at [b]Sang Tao Bat Dang Thuc[/b] (or Secrets in Inequality -Pham Kim Hung) . And this problem is example 3.3.4 (page 257) in that book.

since the inequality is homogene we can suppose that  :  $ abc\equal{}1$
so by AM-GM :

$ \sum\sqrt{a\plus{}3}\,\geq\,3\sqrt[6]{28\plus{}3(ab\plus{}ac\plus{}bc)\plus{}9(a\plus{}b\plus{}c)}\geq3\sqrt[6]{28\plus{}9\plus{}27}\equal{}6$

[quote="mehdi cherif"]since the inequality is homogene we can suppose that  :  $ abc \equal{} 1$
so by AM-GM :

$ \sum\sqrt {a \plus{} 3}\,\geq\,3\sqrt [6]{28 \plus{} 3(ab \plus{} ac \plus{} bc) \plus{} 9(a \plus{} b \plus{} c)}\geq3\sqrt [6]{28 \plus{} 9 \plus{} 27} \equal{} 6$[/quote]

Your proof is so wrong [b]mehdi cherif[/b]
In this case you cannot assume that $ abc \equal{} 1$.
Don't you see that the condition of this ineq is $ ab \plus{} bc \plus{} ca \equal{} 1$.
When you add another condition $ abc\equal{}1$ so now you have two. So check it clearly :wink:

[quote="arqady"][quote="dduclam"]
Arqady,this is [b]Contradiction[/b] method. [/quote]
Wow! Thank you! But it's from which science?  :maybe:
[quote="dduclam"]
If you can't understand the principal of this method,...[/quote]
I don't understand your and Honey_S's reasoning because I think this reasoning is wrong.
[quote="dduclam"]
have a look at [b]Sang Tao Bat Dang Thuc[/b] (or Secrets in Inequality -Pham Kim Hung) . And this problem is example 3.3.4 (page 257) in that book.[/quote]
I can't to take this book. But if it's written in the Hung's  book then Pham Kim Hung was mistaken.  :wink:[/quote]

Arqady,i think you should to lodge the examples for your reasoning  :wink:

[quote="dduclam"]
Arqady,i think you should to lodge the examples for your reasoning  :wink:[/quote]
I don't agree with you!
You need to prove why your method enough for proof of the  original inequality.
The contradiction method works when in an equality
$ p(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)\plus{}q(a^2\plus{}b^2\plus{}c^2)\equal{}1$ we have $ p>0$ and $ q>0.$ :wink:
If $ pq<0$ then our inequality is an example that your reasoning works sometimes.
But why it works always? I think, without proof your reasoning is hallucination.  :lol:  :D  :)

[quote="arqady"][quote="dduclam"]
Arqady,this is [b]Contradiction[/b] method. [/quote]
Wow! Thank you! But it's from which science?  :maybe:
[quote="dduclam"]
If you can't understand the principal of this method,...[/quote]
I don't understand your and Honey_S's reasoning because I think this reasoning is wrong.
[quote="dduclam"]
have a look at [b]Sang Tao Bat Dang Thuc[/b] (or Secrets in Inequality -Pham Kim Hung) . And this problem is example 3.3.4 (page 257) in that book.[/quote]
I can't to take this book. But if it's written in the Hung's  book then Pham Kim Hung was mistaken.  :wink:[/quote]

Sorry Arqady. I heard about this topic in http://www.batdangthuc.net and went here. The method in general is actually not wrong, but for particular problem, we need very small and detailed reasoning to explain clearly what happen.

For this problem,

If $ ab \plus{} bc \plus{} ca \equal{} 3$ then $ \sum \sqrt {a \plus{} 3} \ge 6.$

By CONTRADICTION, assume that $ \sum\sqrt {a \plus{} 3} \le 6$ for a certain triple $ (a,b,c)$, then there exist (by replacing $ a,b,c$ with $ ak,bk,ck$, and $ k > 1$) $ a_1,b_1,c_1$ such that

$ \sum\sqrt {a_1 \plus{} 3} \equal{} 6$, and as a result $ a_1b_1 \plus{} b_1c_1 \plus{} c_1a_1 > 3$.

However, we will prove that: If $ \sum\sqrt {a \plus{} 3} \equal{} 6$ then $ ab \plus{} bc \plus{} ca< 3$ (*).

That is, (*) is a corresponding form of the initial inequality. (*) in a particular meaning is equivalent to the initial one, but to prove it is easier since we can easily remove the square-root.

Let $ x \equal{} \sqrt {a \plus{} 3}$, and similar for $ y,z$, then we have to prove that if $ x \plus{} y \plus{} z \equal{} 3$ then $ \sum (x^2 \minus{} 3)(y^2 \minus{} 3)\le 3$.

The last step is very easy. By this way, we can easily find the best result for $ k$ as you proposed above.

Is this explanation clear enough? I hope so.

Thank you, hungkhtn! You are obviously right!
But I wanted, that dduclam and Honey_S will do it.  :wink: :D :)

Let $ a,b,c>1$ satisfy
\[ a\plus{}b\plus{}c\equal{}\frac1a\plus{}\frac1b\plus{}\frac1c\plus{}6.\]
Prove that
\[ ab\plus{}bc\plus{}ca\leq 9\plus{}6\sqrt2.\]

[quote="Vasc"]The stronger inequality holds for $ ab \plus{} bc \plus{} ca \equal{} 3$:

$ (a \plus{} 3)(b \plus{} 3)(c \plus{} 3) \ge 64$.

These are sharper:

$ (2a \plus{} 3)(2b \plus{} 3)(2c \plus{} 3) \ge 125$;
[/quote]
$uvw$ kills it immediately.

[quote="sqing"]If $ \sqrt {ka^2 \plus{} 1} \plus{} \sqrt {kb^2 \plus{} 1} \plus{} \sqrt {kc^2 \plus{} 1} \le \sqrt {k\plus{}1}\ (a+b+c)$ holds for all positive $ a,$ $ b$ and $ c$ such that $ abc=1$ ,
then$ k_{max}\equal{}?$[/quote]
We can get a best estimation by the following Can's idea:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1975682#p1975682

Thank arqady.
[quote="arqady"] The following nice inequality is true.
Let $a$, $b$, $c$ and $d$ are positive numbers such that $abcd=1.$ Prove that:
\[\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}+\sqrt{8d^2+1}\leq3(a+b+c+d)\][/quote]
Let $a$ and $b$ are positive numbers such that $ab=1.$ Prove that:
\[\sqrt{3a^2+1}+\sqrt{3b^2+1}\leq2(a+b).\]

Nice.
Thanks.
Let $a$ and $b$ are positive numbers such that $ab=1.$ For the positive number $\lambda$ ,prove that\[\sqrt{\lambda a^2+1}+\sqrt{\lambda b^2+1}\leq \sqrt{\lambda+1}(a+b).\]
[b]Proof:[/b] $\sqrt{\lambda a^2+1}+\sqrt{\lambda b^2+1}=\frac{1}{ \sqrt{\lambda+1}}\left(\sqrt{(\lambda+1)a(\lambda a+b)}+\sqrt{(\lambda+1)b(\lambda b+a)}\right)$
$\le \frac{1}{ 2\sqrt{\lambda+1}}\left((\lambda+1)a+(\lambda a+b)+(\lambda+1)b+(\lambda b+a)\right)=\sqrt{\lambda+1}(a+b).$

Let $a$, $b$ and $c$ are positive numbers such that $abc=1.$ Prove that\[\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\leq3(a+b+c).\]

Let $a$, $b$ and $c$ are positive numbers such that $abc=1.$ Prove that\[\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\leq3(a+b+c).\]
We use Minkowsky inequality:
$ \sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\leq \sqrt{8(a+b+c)^2+(1+1+1)^2}=\sqrt{8(a+b+c)^2+9}\leq 3(a+b+c)\iff $
$ 9\leq (a+b+c)^2\iff $
$ 3\leq a+b+c $
This is true because $ 1=abc\leq \frac{(a+b+c)^3}{27} $

Nice.
Thank VadimM.
$ a+b+c\ge 3\sqrt[3]{abc}=3. $

[quote="VadimM"][quote="arqady"][quote="VadimM"]
We use Minkowsky inequality:
$ \sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\leq \sqrt{8(a+b+c)^2+(1+1+1)^2}$
[/quote]
Are you sure?[/quote]
 Yes.
[/quote]
Try $a=1$, $b=2$ and $c=3$. :wink:

Yes. I made a mistake somewhere. Thank you.

[quote=sqing]
Let $a$, $b$ and $c$ are positive numbers such that $abc=1.$ Prove that\[\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\leq3(a+b+c).\][/quote]
Let $a,b,c$ be positive real numbers satisfuing $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$
 [url=https://artofproblemsolving.com/community/c6h1529847p9194660 ]Prove the inequality[/url] \[\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\leq3(a+b+c).\]
[url=https://artofproblemsolving.com/community/c6h597756p3547344]here[/url]


I'm pretty sure it's -2. Since $\sqrt{(-2)}^2 = (-2)^{\frac{2}{2}} = (-2)^1 = -2$.
On second thought, possibly it could be interpreted two ways?

When I put it into my calculator it came up with -2 but also said, "WARNING: Non-real intermediate result".  There was a problem like this on my midterm so I thought there had to be only one answer.

By the order of operations, you take the squareroot first and then square it. Thus, the answer is -2.

Yeah, unless it were $\sqrt{(-2)^2}$. It depends if the exponent is inside or not.

it doesnt work the same way for imaginary numbers (see AoPS chapter 1 if you have it)

[quote="jnsoccer66"][quote="chess64"][quote="jnsoccer66"]...$\sqrt{-2}\times \sqrt{-2}=\sqrt{4}$...[/quote]

...Why? :?[/quote]
 :huh:
$\sqrt{-2} \times \sqrt{-2}=\sqrt{-2\times -2}=\sqrt{4}=2$ just like you would do $\sqrt{xy}\times\sqrt{x^{3}y^{3}}=\sqrt{x\times y\times x^{3}\times y^{3}}=\sqrt{x^{4}y^{4}}=x^{2}y^{2}$.[/quote]

Actually...
$\sqrt{-2}\times \sqrt{-2}=i\sqrt{2}\times i\sqrt{2}=i^2\sqrt{2}^2=-2$, because it's the same thing as $\sqrt{-2}^2$. ;)

jnsoccer, the number under the radical must be $\geq 0$ for that law to work.

Actually, the point is this:

Every complex number (and real numbers are just the special case of complex numbers) have TWO values of the square root. In exponential form, we have $-2=2e^{i(\pi+2n\pi)}$, hence $\sqrt{-2}$ can be either $\sqrt{2}e^{i\pi/2}$ or $\sqrt{2}e^{-i\pi/2}$.

If you multiply SAME values of the root (which is understood as the proper way of squaring), you'll get $2e^{i\pi}=-2$, and if you multiply DIFFERENT values (which is possible, but then it's not squaring), you'll get $2e^{i0}=2.$

Hence, in the "identity" $\sqrt{-2}^2=\underline{\sqrt{-2}\cdot\sqrt{-2}}=\sqrt{(-2)\cdot(-2)}=2$, the two underlined roots are actually different numbers, though written in the same way, and that's similar to the "identity" $\sqrt{4}^2=\sqrt{4}\cdot\sqrt{4}=2\cdot(-2)=-4.$


When viewing a topic, the bottom of the page looks like this:

[img]http://img256.imageshack.us/img256/8023/templj5.png[/img]

Is that supposed to happen? If not in a specific topic, it looks different.

Fixed now.

[hide="Index"]Rules_reply_can[/hide]

[quote="Valentin Vornicu"]Fixed now.

[hide="Index"]Rules_replay_can[/hide][/quote]just FYI, I think you have a typo in that index, should be rules_reply_can. :)

[quote="Peter VDD"]Sorry, I made a typo as well, but I meant you wrote replAy. :) Never mind if it's intended.[/quote]Lol, typo to a typo. Fixed now :D Thanks.


[quote="tranvanluan"]Let $ x;y;z \ge 0$and $ z \plus{} x \plus{} y \equal{} 1$
Prove that
$ x\sqrt {y^2 \plus{} 1} \plus{} y\sqrt {z^2 \plus{} 1} \plus{} z\sqrt {x^2 \plus{} 1}\le \frac {2 \plus{} \sqrt {5}}{4}$[/quote]
try $ x\equal{}0,y\equal{}\frac{1}{3},z\equal{}\frac{2}{3}$.

[quote="tranvanluan"]Let $ x;y;z \ge 0$and $ z \plus{} x \plus{} y \equal{} 1$
Prove that
$ x\sqrt {y^2 \plus{} 1} \plus{} y\sqrt {z^2 \plus{} 1} \plus{} z\sqrt {x^2 \plus{} 1}\le \frac {2 \plus{} \sqrt {5}}{4}$[/quote]
By the way, the following inequality holds
If $ a,b,c$ are the side lengths of a triangle such that $ a\plus{}b\plus{}c\equal{}1$. Then
\[ a\sqrt{b^2\plus{}1}\plus{}b\sqrt{c^2\plus{}1}\plus{}c\sqrt{a^2\plus{}1} \le \frac{2\plus{}\sqrt{5}}{4}\]
:)


Demostrar que para todo número real irracional $ a$, hay números reales irracionales $ b$ y $ c$ tales que $ a\plus{}b$ y $ ac$ son ambos racionales mientras que $ ab$ y $ a\plus{}c$ son ambos irracionales.

Alguna idea :?:

si $ a^2$ es racional dale a $ b\equal{}a^2\minus{}a$, si $ a^2$ es irracional dale a $ b\equal{}\minus{}a$
Sean $ c_1\equal{}1/a$, $ c_2\equal{}\minus{}1/a$, ambos irracionales, notar que $ ac$ es racional. $ a\plus{}c_1\equal{}(a^2\plus{}1)/a$, $ a\plus{}c_2\equal{}(a^2\minus{}1)/a$, la suma de ambos numeros es $ 2a$: irracional, luego alguno de los dos mencionados debe ser irracional.

Bastante interesante, podrías postear las propiedades que se tiene en los números irracionales o talvez algún link  :D


For a positive integer $ n,$ prove that

$ \sum\limits_{k \equal{} 0}^n {\frac{1}{{\dbinom nk}}}  \equal{} \frac{{n \plus{} 1}}{{2^{n \plus{} 1} }}\sum\limits_{k \equal{} 1}^{n \plus{} 1} {\frac{{2^k }}{k}} .$

Define $ a_n\equal{}\sum_{k\equal{}0}^n \frac{2^{n\plus{}1}}{n\plus{}1} \, \frac{1}{{n\choose k}}$ .

$ a_{n\minus{}1}\minus{}a_{n\minus{}2}\equal{}\sum_{k\equal{}0}^{n\minus{}1} \frac{2^n}{n} \, \frac{1}{{n\minus{}1\choose k}}\minus{}\sum_{k\equal{}0}^{n\minus{}2} \frac{2^{n\minus{}1}}{n\minus{}1} \, \frac{1}{{n\minus{}2\choose k}}$

$ \equal{}\frac{2^n}{n}\plus{}\sum_{k\equal{}0}^{n\minus{}2} \underbrace{\left[\frac{2^n}{n} \, \frac{1}{{n\minus{}1\choose k}}\minus{}\frac{2^{n\minus{}1}}{n\minus{}1} \, \frac{1}{{n\minus{}2\choose k}} \right]}_{b_k}$

whereat $ b_k\equal{}\frac{2^{n\minus{}1}}{n\, (n\minus{}1)} \,\frac{n\minus{}2\minus{}2k}{{n\minus{}2 \choose k}}$

Because of $ b_{n\minus{}2\minus{}k}\equal{}\minus{}b_k$ the sum $ \sum_{k\equal{}0}^{n\minus{}2} b_k$ must be $ 0$ .

Hence $ a_n\minus{}\underbrace{a_{\minus{}1}}_{\equal{}0}\equal{}\sum_{k\equal{}1}^{n\plus{}1} \left(a_{k\minus{}1}\minus{}a_{k\minus{}2}\right)\equal{}\sum_{k\equal{}1}^{n\plus{}1}\frac{2^k}{k}$ .


calculus
integration
LaTeX
calculus computations

Let $ I_n = \int\frac {x^n}{1 + x^2}\,dx$. Prove that $ I_n = \frac {x^{n - 1}}{n - 1} - I_{n - 2}$.

Moderator comments:
1) Learn basic $ \text{\LaTeX}$, please!
2) Choose more informative titles for your posts!
3) Use normal fonts, spelling, etc.

[hide]just differentiate both sides of $ I_n \equal{} \frac {x^{n \minus{} 1}}{n \minus{} 1} \minus{} I_{n \minus{} 2}$...[/hide]

Calculate $ I_n \plus{} I_{n \minus{} 2}\ (n>1).$


Idk if I have the right answers to these problems......... :?:

1. Each day, two out of three teams in a class are randomly selected to participate in a MC trial competition. What is the probability htat Team A is selected on on at least two of the next three days?
[hide]My answer was 4/9[/hide]

2. what is the sum of all of hte 4 digit positive integers that can be written with the digits 1,2,3,4 if each digit must be used exactly once in each four digit positive integer?
[hide]I got 66660[/hide]

I don't know about #1, but I do know #2 is correct.

Problem one:
Using complementary counting, we find the chances that team A is NOT selected on at least two of the next three days. The odds of team A being selected on exactly two days is $ 3 \cdot \frac {1}{3} \cdot \frac {1}{3} \cdot \frac {2}{3} \equal{} \frac {2}{9}$. The odds that team A is not selected on all three days is $ \left( \frac {1}{3} \right)^3 \equal{} \frac {1}{27}$. Adding them yields $ \frac {7}{27}$, so the answer is that subtracted from one, which is $ \boxed{\frac {20}{27}}$.

Hmm, nice problem  :) .

Number two is correct.

Oh....thanks!
Btw, the problems were in 03 chapter sprint. :D
(I think I forgot to add something.....lol)


I could not find a pattern. The number decreases and increases in irregular pattern.  :(

i know.....i was trying to solve it for a long time....

Where did you get this sum ? It's pretty interesting....BTW , the numbers do show a pattern ... I'll cross check that and post the answers a little later.  :lol:

A game ?????? :? ............Anyway , I couldn't land on an exact answer so here's something like the fist step towards solving your question -->

Get the differences of each number and it's next -->
The differences go 107 , 79 , 73 , 183 and 137.
Now , write down the differences of the differences -->
They go 28 , 6 , 110 , 46 .
Now , if you observe in a really careful way ,, you'll see that $594 - 566 = 107 - 79 = 28$....The pattern works for a few more numbers in the series , but that 110 seems to make things go out of place a little.........or I'm missing something.

If you didn't understand this post , tell me , I'll explain the stuff with a diagram....

I think the next one is $1279$ but I may have messed up some calculations ...

Haven't found a pattern that I know will work yet.  :(

Hey n0name ,
Here's something that can probably help you understand better -
[code]
594         487       566        493         310          447
     107         79         73          183         137
            28         6           110          46
                  22       104           64
                       82          40
                             42
[/code]

Now see if that makes some sense.BTW , moderators , I think it will be better if you move this to the intermediate topics.....An intermediate student will probably be able to answer this easily...

EDIT : That just doesn't look right once it's posted ....HELP!!!!It's OK in the preview!!!!!!!!!!

P.S: Hey tarquin , could you explain that 1279 ??

uh, noname, are you the one that posted all over on http://www.mathgoodies.com/forums/topic.asp?TOPIC_ID=23418 and http://www.studentquestions.com/homework_help/pg1/42.html ?  :P

no i didnt post those......i think i got the answers....sorry to spoil it, but are the answers 200 and 351?


Let $ P(x)$ be a polynomial with integer coefficients. It is known that $ P(a) \equal{} P(b) \equal{} P(c) \equal{} \minus{}1$, where a, b and c are different integers. Prove that $ P(x)$ does not have integer roots.

$ P(x) \equal{} q(x)(x\minus{}a)(x\minus{}b)(x\minus{}c) \minus{}1$ with $ a,b,c \in \mathbb{Z}$ and $ q(x)$ has integer coefficients

Suppose there is an integer root $ x \equal{} x_0$

$ q(x_0)(x_0\minus{}a)(x_0\minus{}b)(x_0\minus{}c) \equal{} 1$

we have $ q(x_0), (x_0\minus{}a), (x_0\minus{}b), (x_0\minus{}c) \in \mathbb{Z}$ and the values are distinct because $ a,b,c$ are distinct.

But there are no three distinct integers whose product is $ 1$.


Consider $ z_1,z_2,...,z_n \in \mathbb{C}, n \in \mathbb{N}, n \ge 3$ so that:
$ |z_1|\equal{}|z_2\plus{}z_3\plus{}...\plus{}z_n|$
$ |z_2|\equal{}|z_1\plus{}z_3\plus{}...\plus{}z_n|$
...
$ |z_n|\equal{}|z_1\plus{}z_2\plus{}...\plus{}z_{n\minus{}1}|$
Prove that $ z_1\plus{}z_2\plus{}...\plus{}z_n\equal{}0$.

See this problem, attributed to Nazar Agakhanov. [url]http://www.mathlinks.ro/viewtopic.php?t=147169[/url]
Anyways, this type of problem is easier solved using vectors - see solutions above.

[quote="moldovan"]Consider $ z_1,z_2,...,z_n \in \mathbb{C}, n \in \mathbb{N}, n \ge 3$ so that:
$ |z_1| \equal{} |z_2 \plus{} z_3 \plus{} ... \plus{} z_n|$
$ |z_2| \equal{} |z_1 \plus{} z_3 \plus{} ... \plus{} z_n|$
...
$ |z_n| \equal{} |z_1 \plus{} z_2 \plus{} ... \plus{} z_{n \minus{} 1}|$
Prove that $ z_1 \plus{} z_2 \plus{} ... \plus{} z_n \equal{} 0$.[/quote]
Let $ k \equal{} \sum_{i\equal{}0}^n z_i$ Then $ |z_i| \equal{} |k\minus{}z_i| \iff z_i\overline{z_i} \equal{} k\overline{k} \minus{} k\overline{z_i} \minus{} z_i\overline{k} \plus{} z_i\overline{z_i}$ $ \iff k\overline{k} \equal{}  k\overline{z_i} \plus{} z_i\overline{k}$ Summing up we get:
$ n k \overline{k} \equal{} k \sum_{i\equal{}0}^n \overline{z_i} \plus{} \overline{k} \sum_{i\equal{}0}^n z_i \equal{} 2 k \overline{k} \iff$
$ (n\minus{}2) k \overline{k} \equal{} 0 \iff k \overline{k}\equal{}0 \iff |k| \equal{} 0 \iff k \equal{} 0 \iff \sum_{i\equal{}0}^n z_i \equal{} 0$. QED


How do you Tex the similar triangles sign?

??? I was looking for the code. Also, how do you code the arc symbol?

\sim

\stackrel{\frown}{whatever you want the arc to be over}


suppose that:$a \geq b \geq c$ we have $a^2 \geq b^2 \geq c^2$ and
$\frac{1}c \geq \frac{1}b \geq \frac{1}a$ using inequalty [color=blue]tchebychev[/color] we have : $\frac{a^2}b + \frac{b^2}c + \frac{c^2} a \geq \frac{1}3(a^2+b^2+c^2)(\frac{1}c+ \frac{1}b +\frac{1}a)1 =\frac{1}3(a^2+b^2+c^2)(\frac{1}a+ \frac{1}b +\frac{1}c)(a+b+c)$ ,"because $a+b+c=1$"
and $(a+b+c)(\frac{1}a+ \frac{1}b +\frac{1}c) \geq 9$ for $a,b,c>0$

[quote="tchebytchev"]suppose that:$a \geq b \geq c$ we have $a^2 \geq b^2 \geq c^2$[/quote]

I think you can't suppose that since the LHS is cyclic.

From $ab^2(b-c)^2+bc^2(c-a)^2+ca^2(a-b)^2\geq 0$ we have $ab^4+bc^4+ca^4+abc(ab+bc+ac)\geq 2abc(a^2+b^2+c^2)$.

Now it's enough to prove $a^2b^3+b^2c^3+c^2a^3\geq abc(ab+ac+bc)$..

This is trivial after aplaying CBS on vectors $(\frac{\sqrt{b}}c,\frac{\sqrt{c}}a,\frac{\sqrt{a}}b)$ and $(\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}},\frac{1}{\sqrt{a}})$

[quote="dule_00"]Now it's enough to prove $a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3}\geq abc(ab+ac+bc)$.[/quote]

It's Muirhead for $(3,2,0)$ and $(2,2,1).$

[quote="bodan"][quote="dule_00"]Now it's enough to prove $a^2b^3+b^2c^3+c^2a^3\geq abc(ab+ac+bc)$.[/quote]

It's Miurhead for $(3,2,0)$ and $(2,2,1).$[/quote]

It isn't Miurhead because LHS is cyclic.

$\displaystyle\sum a\sum\frac{a^2}b = \sum a^2+\sum\frac{(a+c)a^2}b\ge 3\sum a^2\Leftrightarrow$
$\Leftrightarrow \displaystyle \sum\frac{(a+c)a^2}b\ge 2\sum a^2$
$\displaystyle\sum\frac{a^3}b+\sum ab\ge 2\sum a^2$
Not it is enough to prove that $\displaystyle\sum \frac cba^2\ge\sum ab$
But from CBS$\displaystyle\sum\frac cba^2\sum bc\ge\left(\sum ab\right)^2$ from wich follows the conclusion.

Oh,nice :) .
And trebusep, $a^2\ge b^2\ge c^2$ , $\frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}$.
We can't use it.

Needn't any well-known inequalities, just homogenize, the result follows.
Sorry, I don't have much time.

This inequality can be broken in
$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\geq\ a^2+\ b^2+\ c^2$
                                        and
$\frac{a^2c}{b}+\frac{b^2a}{c}+\frac{c^2b}{a}\geq\ a^2+\ b^2+\ c^2$.
  The first inequality followes immediately by Cauchy and the second it's trivial applying Schur.

I have a proof:
$\displaystyle\sum a\sum\frac{a^2}b = \sum a^2+\sum\frac{(a+c)a^2}b\ge 3\sum a^2$
$\Leftrightarrow \displaystyle \sum\frac{(a+c)a^2}b\ge 2\sum a^2$
apply chauchy ineq$\sum{b(a+c)}\sum\frac{(a+c)a^2}{b^2}\geq (\sum a^2+\sum ab)^2$
so$\sum\frac{(a+c)a^2}{b^2}\geq \frac{(\sum a^2+\sum ab)^2}{2\sum ab}\geq 2\sum a^2$

We denote: $a=\frac{x}{r}$, $b=\frac{y}{r}$, $c=\frac{z}{r}$, with $x,y,z,r>0$.
From condition $a+b+c=1$, results $r=x+y+z$ and $a=\frac{x}{x+y+z}$, $b=\frac{y}{x+y+z}$, $c=\frac{z}{x+y+z}$.
The inequality given is equivalent with:
$(x+y+z)(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x})\geq 3(x^2+y^2+z^2)$ $\Leftrightarrow$
$(*)$ $x^2(\frac{x}{y}+\frac{z}{y})+y^2(\frac{y}{z}+\frac{x}{z})+z^2(\frac{z}{x}+\frac{y}{x})\geq 2(x^2+y^2+z^2)$ $\Leftrightarrow$
Using  $\frac{x}{y}\geq 2-\frac{y}{x}, \forall x,y>0$ and analogues, we obtain:
$(*)$ $\Leftrightarrow$ $\frac{x^2 z}{y}+\frac{y^2 x}{z}+\frac{z^2 y}{x}\geq xy+yz+zx$.
The last one is obvious from Cauchy-Schwarz.

[quote="zhaobin"][quote="Utkirstudios"]view in  solution in jpg[/quote]
Nice solution :)[/quote]
I agree

[quote="Lovasz"]One-line proof:
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3(a^2+b^2+c^2) \\ \iff \sum_{cyc} (a-b)^2 \cdot \frac{c+a}{b(b+c+a)} \ge 0.  \][/quote]
Nice solution!

I came up with a solution that I am not 100% sure is not announced, but I considerred it ‘not too standard’ so here it is:  (please somebody change it to LateX)
The inequality is equivalent to
(a^2/b)+(b^2/c)+(c^2/a)>=3(a^2+b^2+c^2)/(a+b+c)_________ (1) which is homogene so we may suppose abc=1
After equilizing denominators and making cross product it reduces to
a(b^4)+b(c^4)+c(a^4)+(a^2)(b^3)+(b^2)(c^3)+(c^2)(a^3)>=2(a^2+b^2+c^2)_____________(2)
According to CS
 (a(b^4)+b(c^4)+c(a^4))(1/a+1/b+1/c)>=((a^2+b^2+c^2)^2)
(a(b^4)+b(c^4)+c(a^4)>= ((a^2+b^2+c^2)^2)/( (1/a+1/b+1/c))
and
(a^2)(b^3)+ab+ac>=3ab according to AM-GM
summing inequalities of this type gives
(a^2)(b^3)+(b^2)(c^3)+(c^2)(a^3)>=ab+ac+bc
Now (2) reduces to
(((a^2+b^2+c^2)^2)/( (1/a+1/b+1/c)))+ ab+ac+bc>=2(a^2+b^2+c^2)
which is simply AM-GM

Here is my idea:
Set $ab=x^{2}.bc=y^{2},ac=z^{2}$ therefore $a=\frac{xy}z,b=\frac{xz}y,c=\frac{yz}x$ and $(a+b+c)^{2}=1=a^{2}+b^{2}+c^{2}+2(x^{2}+y^{2}+z^{2})$ thus $a^{2}+b^{2}+c^{2}=1-2(x^{2}+y^{2}+z^{2})$ therefore the ineq desired is rewritten as:
$x\frac{y^{3}}{z^{3}}+y\frac{z^{3}}{x^{3}}+z\frac{x^{3}}{y^{3}}\geq 3-6(x^{2}+y^{2}+z^{2})$ which is rewritten as
$\frac{xy^{3}}{z^{3}}+\frac{yz^{3}}{x^{3}}+\frac{zx^{3}}{y^{3}}+6(x^{2}+y^{2}+z^{2})\geq 3$ which is equivalent to
$\sum_{cyc}x^{4}y^{6}+6\sum_{cyc}x^{5}y^{3}z^{3}\geq 3x^{3}y^{3}z^{3}$.For the moment my brain is stuck at this one[probably because I am a bit tired].If anyone cares to prove the last one,be my guest.
PS:It must be easy,because I haven`t used any inequality so far..Good luck

mentioned in 4th post,

what is CBS inequality?

It`s Cauchy-Schwartz
B stands for Buniakovski

You can use just "simple" AM-GM :wink:

Homogenising gives $\sum\left(a^{4}c+a^{2}b^{3}\right)\geq 2\sum a^{3}bc$

Since $a^{4}c+\frac{4}{7}a^{2}b^{3}+\frac{2}{7}c^{2}a^{3}+\frac{1}{7}b^{2}c^{3}\geq a^{4}c+a^{2}b^{2}c \geq 2a^{3}bc$, we are done

[quote="Makaveli"]I came up with a solution that I am not 100% sure is not announced, but I considerred it ‘not too standard’ so here it is:  (please somebody change it to \LaTeX)
The inequality is equivalent to \[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq 3\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\;\;\;\;\; (1)\] which is homogene so we may suppose $abc=1.$

After equilizing denominators and making cross product it reduces to \[ab^{4}+bc^{4}+ca^{4}+a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3}\geq2(a^{2}+b^{2}+c^{2}).\;\;\;\;\;(2)\] According to CS \[(ab^{4}+bc^{4}+ca^{4})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq(a^{2}+b^{2}+c^{2})^{2}(ab^{4}+bc^{4}+ca^{4})\geq\frac{(a^{2}+b^{2}+c^{2})^{2}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\] and \[a^{2}b^{3}+ab+ac\geq3ab\] according to AM-GM.
Summing inequalities of this type gives \[a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3}\geq ab+ac+bc.\] Now (2) reduces to \[\frac{(a^{2}+b^{2}+c^{2})^{2}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+ab+ac+bc\geq 2(a^{2}+b^{2}+c^{2})\] which is simply AM-GM[/quote]

is 3 the largest constant? how we prove it?

[quote="red3"]$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-3\frac{a^2+b^2+c^2}{a+b+c}=\frac{1}{abc(a+b+c)(ab+bc+ca)}((a^3+abc)(b^2-ac)^2+(b^3+abc)(c^2-ab)^2+(c^3+abc)(a^2-bc)^2)$

also one line :oops:[/quote]

is also true because
$LHS = \sum{(a-b)^2(\frac{1}{b}-1)}$
and $a,b,c \le 1$ (because $a+b+c = 1$ and $a,b,c \in R^+$ )

[quote=hungkhtn]I remember that this problem was posted in this forum. That is

\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{3(a^2+b^2+c^2)}{a+b+c}. \]
(Even it was expended). Solution by S.O.S as Lovasz is the simplest one.[/quote]
Let $a,b,c>0$.[url=https://artofproblemsolving.com/community/c6h1350736p7374983]Prove:[/url]
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{2(a^2+b^2+c^2)}{a+b+c}+\sqrt[6]{\frac{a^6+b^6+c^6}{3}} $$

[quote=hungkhtn]I remember that this problem was posted in this forum. That is

\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3\frac{a^2+b^2+c^2}{a+b+c}. \]
(Even it was expended). Solution by S.O.S as Lovasz is the simplest one.[/quote]

<=>
$x,y,z>0,$
\[M={x}^{4}z-2\,{x}^{3}zy+{z}^{2}{x}^{3}+{x}^{2}{y}^{3}+x{y}^{4}-2\,x{y}^{
3}z-2\,y{z}^{3}x+{y}^{2}{z}^{3}+y{z}^{4}\geq 0\]

$x$=min($x,y,z$),then

\[M=z \left( {z}^{2}+yz+2\,zx-xy+{x}^{2} \right)  \left( x-y \right)
 \left( x-z \right) +x \left( {y}^{2}+xy+zx+{z}^{2} \right)  \left( y-
z \right) ^{2}\geq 0\]

easy do.

[quote=Valentin Vornicu]Let $a,b,c>0$ be real numbers with sum 1. Prove that \[ \frac{a^2}b + \frac{b^2}c + \frac{c^2} a \geq 3(a^2+b^2+c^2) . \][/quote]
[url=https://artofproblemsolving.com/community/c4h1751026p11415319]NaPrai:[/url]
\begin{align*}LHS \ge RHS &\Leftrightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-1\ge 3(a^2+b^2+c^2)-1\\&\Leftrightarrow \left(\frac{a^2}{b}-2a+b\right)+\left(\frac{b^2}{c}-2b+c\right)+\left(\frac{c^2}{a}-2c+a\right)\ge 3(a^2+b^2+c^2)-(a+b+c)^2 \\&\Leftrightarrow \frac{1}{b}(a-b)^2+\frac{1}{c}(b-c)^2+\frac{1}{a}(c-a)^2 \ge (a-b)^2+(b-c)^2+(c-a)^2\\&\Leftrightarrow \left(\frac{1}{b}-1\right)(a-b)^2+\left(\frac{1}{c}-1\right)(b-c)^2+\left(\frac{1}{a}-1\right)(c-a)^2\ge 0\end{align*}which is obviously true because $a,b,c<1$
[url=https://artofproblemsolving.com/community/c6h1631279p10241183]here[/url]
[url=https://artofproblemsolving.com/community/c6h435658p2458309]here[/url]

[quote=hungkhtn]I remember that this problem was posted in this forum. That is

\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{3(a^2+b^2+c^2)}{a+b+c}. \]
(Even it was expended). Solution by S.O.S as Lovasz is the simplest one.[/quote]

Let $a,b,c$ be positive real numbers . Prove that
$$\frac{b}{a^2}+\frac{c}{b^2}+\frac{a} {c^2}\ge \frac{3(a+b+c)}{a^2+b^2+c^2},$$
ܽ$$ \frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge \frac{3(a^3+b^3+c^3)}{a+b+c}. $$
(Nguyen Viet Hung)

Let $a,b,c$ be positive real numbers . For  $2\leq n\in N,$ prove or disprove ܽ$$ \frac{a^n}{b}+\frac{b^n}{c}+\frac{c^n}{a} \ge \frac{3(a^n+b^n+c^n)}{a+b+c},$$
$$ \frac{b}{a^n}+\frac{c}{b^n}+\frac{a}{c^n}\ge \frac{3(a+b+c)}{a^n+b^n+c^n}. $$

Let $a,b,c$ be positive real numbers .[url=https://artofproblemsolving.com/community/c6h2030698p15143756]Prove that[/url]
\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geqslant a+b+c+\frac{(a-b)^2}{a+\frac{6}{25}b}.\]
\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geqslant a+b+c+\frac{(a-b)^2}{a+kb}.\]where $k_{\min}=0.2133885325$
[url=https://artofproblemsolving.com/community/c6h566836p3319686]h[/url]


inequalities
algebra
polynomial
inequalities unsolved

Prove the inequality:

1/(1-xy) + 1/(1-yz) + 1/(1-zx) <= 27/8
when x,y,z are positive real numbers such that x+y+z = 1

Please help me solve this as many way as possible
Thank a lot. :)

The inequality is: $ \sum\frac{1}{1\minus{}xy} \leq \frac{27}{8}$
$ \Leftrightarrow \sum (1\minus{}xy)(1\minus{}yz) \leq \frac{27}{8}(1\minus{}xy)(1\minus{}yz)(1\minus{}zx)$
$ \Leftrightarrow 8\sum (1\minus{}xy\minus{}yz\plus{}xy^2z) \leq 27(1\minus{}(xy\plus{}yz\plus{}zx)\plus{}xyz(x\plus{}y\plus{}z)\minus{}(xyz)^2)$
$ \Leftrightarrow 24\minus{}16(xy\plus{}yz\plus{}zx)\plus{}8xyz \leq 27\minus{}27(xy\plus{}yz\plus{}zx)\plus{}27xyz\minus{}27(xyz)^2$
$ \Leftrightarrow 11(xy\plus{}yz\plus{}zx)\plus{}27(xyz)^2\leq 19xyz\plus{}3$

From AM-GM inequality, we obtain:
$ xyz\leq\frac{(x\plus{}y\plus{}z)^3}{27}\equal{}\frac{1}{27}\Rightarrow 27(xyz)^2\leq\frac{1}{27}$
$ \Rightarrow LHS\leq 11(xy\plus{}yz\plus{}zx)\plus{}\frac{1}{27}$

Now we have to prove:
$ 11(xy\plus{}yz\plus{}zx)\leq 19xyz\plus{}\frac{80}{27}$
$ \Leftrightarrow 11(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)\leq 19xyz\plus{}\frac{80}{27}(x\plus{}y\plus{}z)^3$

Put $ f(x,y,z)\equal{}19xyz\plus{}\frac{80}{27}(x\plus{}y\plus{}z)^3\minus{}11(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)$ we have:
$ f(0;0;0),f(1;0;0),f(1;1;0)\geq 0$, which ends the proof.

Sorry, but I don't get it:

[quote="Mashimaru"]
Put $ f(x,y,z) \equal{} 19xyz \plus{} \frac {80}{27}(x \plus{} y \plus{} z)^3 \minus{} 11(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)$ we have:
$ f(0;0;0),f(1;0;0),f(1;1;0)\geq 0$, which ends the proof.[/quote]

Why $ f(0;0;0),f(1;0;0),f(1,1,0)\geq 0$ end the proof?
Is there another way to prove
$ 19xyz \plus{} \frac {80}{27}(x \plus{} y \plus{} z)^3 \minus{} 11(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \geq 0$?

[quote="Mashimaru"]The inequality is: $ \sum\frac {1}{1 \minus{} xy} \leq \frac {27}{8}$
$ \Leftrightarrow \sum (1 \minus{} xy)(1 \minus{} yz) \leq \frac {27}{8}(1 \minus{} xy)(1 \minus{} yz)(1 \minus{} zx)$
$ \Leftrightarrow 8\sum (1 \minus{} xy \minus{} yz \plus{} xy^2z) \leq 27(1 \minus{} (xy \plus{} yz \plus{} zx) \plus{} xyz(x \plus{} y \plus{} z) \minus{} (xyz)^2)$
$ \Leftrightarrow 24 \minus{} 16(xy \plus{} yz \plus{} zx) \plus{} 8xyz \leq 27 \minus{} 27(xy \plus{} yz \plus{} zx) \plus{} 27xyz \minus{} 27(xyz)^2$
$ \Leftrightarrow 11(xy \plus{} yz \plus{} zx) \plus{} 27(xyz)^2\leq 19xyz \plus{} 3 ( 1 )$

[/quote]
Let $ ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r$. We have :

$ ( 1 ) \leftrightarrow 11q \plus{} 27r^2 \le\ 19r \plus{} 3$

Follow Schur : $ r \ge\ \frac {4q \minus{} 1}{9} \rightarrow q \le\ \frac {9r \plus{} 1}{4}$

Need prove  $ \frac {11(9r \plus{} 1)}{4} \plus{} 27r^2 \minus{} 19r \minus{} 3 \le\ 0$

$ \leftrightarrow (4r \plus{} 1)(27r \minus{} 1) \le\ 0$

Which is obvious! :wink:

We have done! :)

:rotfl:
Now I've done. We have:
$ 11(xy \plus{} yz \plus{} zx)\leq 19xyz \plus{} \frac {80}{27}$
$ \Leftrightarrow yz(19x \minus{} 11) \plus{} 11x^2 \minus{} 11x \plus{} \frac {80}{27} \ge 0$
$ Put f(u) \equal{} u(19x \minus{} 11) \plus{} 11x^2 \minus{} 11x \plus{} \frac {80}{27}\ge 0$

$ x \equal{} \frac {11}{19}, f(u) > 0$
$ x < > \frac {11}{19}. We have: 0 < yz \leq \frac {(y \plus{} x)^2}{4} \equal{} \frac {(1 \minus{} x)^2}{4}$
$ f(0) > 0$
$ f(\frac {(1 \minus{} x)^2}{4}) \equal{} \frac {19(x \plus{} \frac {23}{57})(x \minus{} \frac {1}{3})^2}{4}\ge 0$
That ends the proof. :blush:
Great thank to Mashimaru and onlylove_math.

You are right, Mashimaru is also(almost!).
Since  $ f$ is homogeneous symmetric poynomial with dgree $ 3$ and $ f(1,1,1), f(1,1,0), f(1,0,0) \geq 0$ we are done.


Source: Elias Saab

A garage door opener has a  ten-digit  keypad. Codes to open the door must consist of  4  digits with no adjacent digits the same. How many codes are possible?

could this be counted as one of possibilities?

1,3,5,7

and

3,1,5,7?

hm...
then I have no idea how to solve this problem.....

It would take me at least a day....

I;m not really good at finding like possible numbers and infinities like higher leveled math...

[hide="hint"]Think about how many different choices there are for the first digit.  Then, with that digit in place, how many choices will there be for the second digit, etc...[/hide]

[hide]We have 10 choices for the first, and 9 for the next, since we can't duplicate the first, 9 for the next, since it can't be the second but it can be the first since first and third are not adjacent, and 9 for the last, so it isn't the same as the third.

So we have 810*9 choices.[/hide]

Just a question:

Does the keypad look like:

[code]
1 2 3 4 5 6 7 8 9 0
[/code]

pr

[code]
1 2 3
4 5 6
7 8 9
  0
[/code]

Treething -- it doesn't matter what the keypad looks like.  "Adjacent" in the problem only refers to the order of the digits in the combination, not on the keypad.

A+ -- yes


When I'm writing linguistics papers, I need to write the text of the paper in English, but include tables with examples in different languages.  How do you set the language for an entire table?  Using the \setlanguage{N} command at the start of the table appears to only affect the first item.  For instance, in the following example, only the first $ \tau\iota\mu\acute{\eta}$ is displayed in Greek.

[quote]
\documentclass[11pt]{amsart}
\usepackage{geometry}                % See geometry.pdf to learn the layout options. There are lots.
\geometry{letterpaper}                   % ... or a4paper or a5paper or ...
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\usepackage[polutonikogreek,english]{babel}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}

\begin{document}
\begin{tabular}{ l l l l }
\selectlanguage{polutonikogreek}
  tim'h & tim'h & tim$ \acute{\bar{\alpha}}$ & \selectlanguage{english}`honor'\\
\end{tabular}

\end{document}
[/quote]

The table has already started before you changed the language so cannot apply to the whole table. So move \selectlanguage to be before \begin{tabular}.
If you don't want to have to change the language back again after the table group the whole lot inside {...}
{\selectlanguage ... \end{tabular}}


When solving another problem, I think about a new problem but I haven't found the answer yet.
It is: Find all the pair $(m, n)$ such that: $\large{2^n+3 | 2^m+1}$.
The second question: Is there infinitely many n such that $\large{2^n+1}$ is a prime ?

Your second question is the fabled merenese prime conjecture.

[quote="ZetaX"]No, the second question is the questen iff there are infinetely many Fermat primes (those of type $2^n+1$; it can be shown that $n=2^m$ is a power of two itself already).
The Mersenne primes are those of type $2^n-1$ and then $n$ is necessary prim itself.[/quote]That´s right ZetaX...

My bad, so it is. I dunno, whenever i see 'merennese prime' written somwhere, i immediatly think of $2^n \pm 1$ for some reason.

How about the first question.
I think that there are infinitely many pair (m,n) but I can find all sort of them.


calculus
integration
number theory proposed
number theory

Find all integral solutions to (m2 - n2)2 = 1 + 16n    :D thanks for solve it :D

1.: Don't double-post. (http://www.mathlinks.ro/Forum/viewtopic.php?t=63426 )
2.: Do what ehsan2004 says in the other topic.
3.: Did you just copy it from somewhere¿ It looks like...


algorithm
combinatorics unsolved
combinatorics

Let $ p$, $ m$, $ n$ be three nonnegative reals. Prove that
$ \sum_{k \equal{} 0}^n \Large{n\choose k}\Large{m\choose k}\Large{p \plus{} k\choose m \plus{} n}\equal{}\Large{p\choose n}\Large{p\choose m}$.

This is a Pfaff-Saalschütz summation as in the binomial identity with four variables thread and everything said there holds here, too.

If you don't want to use algorithms for some ideological reason, you can use Vandermonde summation and $\binom nk \binom ki = \binom ni \binom{n-i}{k-i}$ to prove it like this:

\[ \sum_{k=0}^n \binom n k \binom m k \binom {p+k}{ m+n}\\ \&= \sum_{k=0}^n \binom n k \binom m k \sum_{i=0}^{\infty} \binom{p}{m+i}\binom {k}{ n-i}\\ \&= \sum_{k=0}^n \sum_{i=0}^{\infty} \binom m k \binom{p}{m+i}\binom n k\binom {k}{ n-i}\\ \&= \sum_{i=0}^{\infty} \binom{p}{m+i} \sum_{k=0}^n \binom m k \binom n {n-i} \binom {n-(n-i)}{ k-(n-i)}\\ \&=\sum_{i=0}^{\infty} \binom{p}{m+i} \binom n i\sum_{k=0}^n \binom m k \binom {i}{ n-k}\\ \&=\sum_{i=0}^{\infty}\binom n i \binom{p}{m+i} \binom {m+i} n \\ \&=\sum_{i=0}^{\infty}\binom n i \binom{p}{n} \binom{p-n} {m+i- n} \\ \&=\binom{p}{n} \sum_{i=0}^{\infty}\binom n i \binom{p-n} {p-m-i} \\ \&=\binom{p}{n} \binom{p} {p-m} =\binom{p}{n} \binom{p} {m}  \]


Imagine ducks as points in a plane. Three ducks are said to be in a row if a straight line passes through all three ducks. Three ducks, Huey, Dewey, and Louie, each waddle along a different straight line in the plane, each at his own constant speed. Although their paths may cross, the ducks never bump into each other. Prove: If at three separate times the ducks are in a row, then they are always in a row.

[hide="Approach 1"] Parameterize each duck's path.  You should get a solvable system. [/hide]
[hide="Approach 2"] If the ducks are in a row three separate times, consider the intersection(s) of the lines passing through all three ducks at those times. [/hide]


calculus
integration
conics
parabola
calculus computations

Let $ K$ be the curved surface obtained by rotating the parabola $ y \equal{} \frac {3}{4} \minus{} x^2$ about the $ y$-axis.Cut $ K$ by the plane $ H$ passing through the origin and forming angle $ 45^\circ$ for the axis. Find the volume of the solid surrounded by $ K$ and $ H.$

Note that you are not allowed to use Double Integral for the problem.

$ \int_{y \equal{} \minus{}1}^{1} \int_{x \equal{} \minus{} \frac {1}{2} \minus{} \sqrt {1 \minus{} y^2}}^{ \minus{} \frac {1}{2} \plus{} \sqrt {1 \minus{} y^2}} \left(\frac {3}{4} \minus{} (x^2 \plus{} y^2) \minus{} x\right) \, dx \, dy \equal{} \frac {\pi}{2}$


Find the smallest integer $ n\geq 2008$ such that all the coefficients of the expansion of $ (x\plus{}y)^n$ are odd.

the answer is $ 2047$
that is the smallest $ n$ for which $ 2^n\minus{}1>2008$

hint : if  the binary form of n contains a zero then one of the coefficients are even (    $ \binom{n}{i}$  where i has a 1 exactly where n has a zero in it's
binary form.)

[quote="kai_desmond11"]
hint : if  the binary form of n contains a zero then one of the coefficients are even (    $ \binom{n}{i}$  where i has a 1 exactly where n has a zero in it's
binary form.)[/quote]

Eh, I can't really see why...


geometry
number theory proposed
number theory

hi,
assume that "a" is an integer and "p" is a prime number
and (a,p)=1
and "n" is an integer
find such n that we have
(a^n)-n*p=p!*k               (k is an integer)
 :lol:

First of all, this post should be moved, since it's not a geometry problem. Secondly, are you sure the problem is right, because the only solution is $ n\equal{}0$ as otherwise $ p|a$ which is a contradiction to $ (a,p)\equal{}1$.


A robotic arm, suspended over a bin of $2007$ balls, is programmed to read instructions encoded as a sequence of $0$s and $1$s.  A $0$ followed immediately by a $1$ in the sequence causees the arm to add a ball to the bin, while a $1$ followed by a $0$ causes the arm to remove a ball from the bin.  A $1$ followed by a $1$ or a $0$ followed by a $0$ causes no action by the arm.  So, if three numbers $x, y, z$ are fed into the machine in order, it will process $xy$, then $yz$.

(For example, $110$ will do nothing ($11$), and then remove a ball $(10)$.)

After a single [i]finite[/i] instruction sequence, what is the largest number of balls that can remain in the bin?

[hide="Is it just?"]

2008?

Suppose we sit at the side watching, and every time the arm adds a ball, we write down A. Every time the arm removes a ball, we write down R. Then, if after a finite sequence, there are $>2008$ balls, at some point in the sequence of A's and R's we have written down, there are two consecutive A's.
But this is impossible since after we have $01$ in the sequence, we can't have a $01$ again until after there is a $10$, however many $1$'s there are in the intervening period...

The finite sequence $01$ achieves this maximum.

(That was kind of long-winded, but I kept feeling I was making a mistake)[/hide]

Yes. :)

[hide="In other words..."] Divide up the sequence into blocks of consecutive $1$s or $0$s.  Every time we switch from $0$s to $1$s, we add a ball, and every time we switch from $1$s to $0$s, we remove a ball.

Of course, depending on the first number in the sequence, every time we switch, we must switch back.

In other other words, imagine we are traveling along one of two tracks - a $0$ track and a $1$ track.  We begin along one of the tracks, and every time we switch to the other track, we either continue to proceed or we switch back to the first track.  In the end, we can only move a total of one track in either direction, which corresponds to either one addition or one removal of a ball. [/hide]


Let $ T$ be the centroid of $ \triangle{ABC}$and let $ TM,TN,TP$ be the altitudes in the triangles $ ATB,BTC,TCA,$respectively.
Show that the area of $ \triangle{ABC}$ is at least four times larger than the area of $ \triangle{MNP}$


The circle $ w\equal{}C(I)$ touches the sides of $ \triangle ABC$ in $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ .
Denote the point $ P\in DE$ for which $ PC\perp BI$ . Prove that the point $ P$ belongs
to the common cord between the circle $ w$ and the circle with diameter $ [BC]$ .

Call H the feet of the perpendicolar from B to CP.
We have to proove that P stay on the radical axes of the circle with diameter BC and $ w$ and so that $ \overline{PE} \cdot \overline{PD} \equal{} \overline{PH} \cdot \overline{PC}$.
But C, D, I, E, H stay on the same circle $ \Gamma$ with diameter CI cause $ \angle CHI \equal{} \angle CEI \equal{} \angle CDI \equal{} \frac{\pi}{2}$.
Then for the power of P w.r.t. $ \Gamma$ we have $ \overline{PE} \cdot \overline{PD} \equal{} \overline{PH} \cdot \overline{PC}$.


Classical inequality, clasical solution(this means an ugly solution :) )! :D

(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)>=(ab+bc+ca)^3
we divide it by a^2b^2c^2 and we denote a/b=x, c/a=z and b/c=y and the inequality is equivalent to:
(1+x+1/x)(1+y+1/y)(1+z+1/z)>=(1+z+1/y)(1+x+1/z)(1+y+1/x)
we can see that
(1+z+1/y)(1+x+1/z)(1+y+1/x)<=(1+x+1/y)(1+z+1/z)(1+y+1/x)<=(1+x+1/x)(1+y+1/y)(1+z+1/z)
which is easy to prove, you get that these inequalities are true iff (z-x)(z-y)>=0 which is true if we consider z>=y>=x (the ineq is simetrical). we have equality iff a=b=c.

cheers! :D :D


Let $m$ be a positive integer. Define the sequence $a_{0}, a_{1}, a_{2}, ...$ by $a_{0}=0$, $a_{1}=m$ and $a_{n+1}=m^{2}a_{n}-a_{n-1}$ for $n=1,2,3 ...$. Prove that an ordered pair $(a,b)$ of non-negative integers, with $a \leq b$, gives a solution to the equation
$\frac{a^{2}+b^{2}}{ab+1}=m^{2}$
if and only if $(a,b)$ is of the form $(a_{n},a_{n+1})$ for some $n \geq 0$.

Thanks. Would a mod like to delete this topic?


Help me quickly with this math problem:
Let $ a$,$ b$,$ c>0$ such that $ xy+yz+zx=1$
$ \sum_{cyc}\sqrt{\frac{(1+y^2)(1+z^2)}{1+x^2}}$=[size=150][b]?[/b][/size]
If you can't solve it, just give me your suggestion.

Hint : $ \triangle ABC \ : \  cotg A\equal{}x\  ;\  cotg B\equal{}y \ ;\ cotg C\equal{}z$

After 5 minutes, very fast and nice.
Thank you hophinhan :lol:
But I wonder if any one has another proof not using $ sin,cos,tan,cotg$

[quote="Unidranix"]Help me quickly with this math problem:
Let $ a$,$ b$,$ c > 0$ such that $ xy + yz + zx = 1$
$ \sum_{cyc}\sqrt {\frac {(1 + y^2)(1 + z^2)}{1 + x^2}}$=[size=150][b]?[/b][/size]
If you can't solve it, just give me your suggestion.[/quote]
Do you want a number??
Cause i found $ 2(x+y+z)$ :huh:

I can't believe that this problem is too easy.
Sorry because of wasting your time.

$ x,y,z \in R^ \plus{}$ such as $ (x\plus{}y\plus{}z)\equal{}xyz$.

Calculate  :  $ \sum \frac {\sqrt {(1 \plus{} x^2)(1 \plus{} y^2)} \minus{} \sqrt {1 \plus{} x^2} \minus{} \sqrt {1 \plus{} y^2}}{xy}$

@: now, could someone give me a solution without  trigonometry . :P  :P

Yo, that problem is quite simple
We have $ \sum_{cyc}(\frac{1}{xy})\equal{}1$
Then $ 1\plus{}\frac{1}{x^2}\equal{}(\frac{1}{x}\plus{}\frac{1}{y})(\frac{1}{x}\plus{}\frac{1}{z})$
After expanding, the result is 0  :lol:
[hide]Hey, Nhạn, nay cũng lên ML àh  :lol: [/hide]


advanced fields
advanced fields theorems

Hi all,

here is one more problem to think of...thanks for your help!!

Suppose a metal rod loses heat across the letaral boundary according to the equation:

[img]http://www.yu-alumni.com/futurebound/1.jpg[/img]                            ; 0<x<1

and suppose we keep the ends of the rod at U(0,t)=1
                                                                U(1,t)=1

Find the steady- state temperature of the rod. Where is the heat flowingin this problem?[/img]

Just recall that the steady state is the one with $u_{t}=0$ and solve the corresponding boundary problem for the second order ODE you get. As to the heat flow, assuming that $\beta>0$, your equation descrides an iron rod frozen in a block of ice except the ends, which stick out of the ice and which you hold with your warm hands. Now guess what will happen.


if tanA tan(A+B)=k
prove that (k+1)cos(2A+B)=(1-k)cosB

\begin{eqnarray} \frac{\cos (2A + B)}{\cos B} &=& \frac{1 - k}{1 + k} \nonumber\\ \cos 2A - \sin 2A\tan B &=& \frac{1 - k}{1 + k} \nonumber \end{eqnarray}

From the first we get:
\begin{eqnarray} \frac{1 - k}{1 + k} &=& \frac{1 - 2\tan A\tan B - \tan^2 A}{1 + \tan^2 A} \nonumber \\ &=& \frac{1 - \frac{\sin^2 A}{\cos^2 A} - 2\frac{\sin A\sin B}{\cos A\cos B}}{\frac{1}{\cos^2 A}} \nonumber \\ &=& \cos^2 A - \sin^2 A - 2\sin A\cos A\frac{\sin B}{\cos B} \nonumber \\ &=& \cos 2A - \sin 2A\tan B \nonumber \end{eqnarray}
So it's proved.

Because:(k+1)(1-k)=(k+1)[1-tan(A+B)tanA]=(1-k)[1+tan(A+B)tanA]
              Multiply cos(A+B)cosA to the formular above we can get :(By using sinA=tanA*cosA)
             (k+1)[cos(A+B)cosA -sin(A+B)sinA]=(1-k)[cos(A+B)cosA +sin(A+B)sinA]
     By using the formular: cos(A+/-b)=cosAcosB-/+sinAsinB we can get :
                               (k+1)cos(2A+B)=(1-k)cosB
    That's what we have proved!


induction
quadratics
function
geometry
parallelogram
combinatorics proposed
combinatorics

Given a set $ \mathcal{H}$ of points in the plane, $ P$ is called an "intersection point of $ \mathcal{H}$" if distinct points $ A,B,C,D$ exist in $ \mathcal{H}$ such that lines $ AB$ and $ CD$ are distinct and intersect in $ P$.
Given a finite set $ \mathcal{A}_{0}$ of points in the plane, a sequence of sets is defined as follows: for any $ j\geq0$, $ \mathcal{A}_{j+1}$ is the union of $ \mathcal{A}_{j}$ and the intersection points of $ \mathcal{A}_{j}$.
Prove that, if the union of all the sets in the sequence is finite, then $ \mathcal{A}_{i}=\mathcal{A}_{1}$ for any $ i\geq1$.

Induction on the number of prime divisors of $ n$ (not elegant, but it works).

As for the induction start assume $ n\equal{}p^{t}$. If $ p\equal{}2$ holds, set $ a\equal{}1,b\equal{}k\minus{}1$ and remember that $ k$ must be even then. If $ p>2$ choose any residue $ a$ different from $ 0,k\bmod{p}$ and set $ b\equal{}k\minus{}a$.

As for the induction step write $ n\equal{}p^{t}n'$ with $ p$ an odd prime and $ n'$ not divisble by $ p$. Since $ n,n'$ have the same parity, we can assume the existence of integers $ a'\plus{}b'\equal{}k$, which are both coprime to $ n'$. Consider the quadratic function $ x\rightarrow f(x)\equal{}(a'\plus{}xn')(b'\minus{}xn')$, which is not identical to zero modulo $ p$, since $ n'^{2}$ is not divisible by $ p$. Therefore, there exists $ x$ such that $ f(x)$ is not divisible by $ p$ and we can set $ a\equal{}a'\plus{}xn', b\equal{}b'\minus{}xn'$ to complete the proof.

Suppose $ \cup_{j\equal{}0}^{\infty}$ is finite. Then for some $ j$, $ A_j\equal{}A_{j\plus{}1}\equal{}\ldots$ and $ A_j$ is finite. Suppose in $ A_j$ we have $ 4$ points such that one of them is strictly inside the nondegenerate triangle formed by the other $ 3$. Among all such quadruples of points (in finite number, because $ A_j$ is finite), take $ 4$ such that the area of the triangle containing the $ 4$th point is minimal. Say these points are $ P\in\textrm{Int}(ABC)$. Then $ A'\equal{}AP\cap BC$, $ B'\equal{}BP\cap CA$ and $ C'\equal{}CP\cap AB$ are all in $ A_{j\plus{}1}$ and hence, by the choice of $ j$, in $ A_j$. It is easy to see that $ P\in\textrm{Int}(A'B'C')$ and $ [A'B'C']<[ABC]$. Then the choice of $ P,A,B,C$ is contradicted. Thus no such points exist.
Assume there are $ 4$ points $ A,B,C,D$ in $ A_j$ such that $ ABCD$ is a non-degenerate 'normal' convex quadrilateral. Let $ P\equal{}AC\cap BD$. Then $ P\in A_{j\plus{}1}$ and hence $ P\in A_j$. Assume $ ABCD$ is not a parallelogram. Assume $ AB$ isn't parallel to $ CD$. Then $ Q\equal{}AB\cap CD$ is in $ A_j$ and from the convexity of $ ABCD$, we have $ P\in\textrm{Int}(QAB)$ or $ P\in\textrm{Int}(QCD)$. Contradiction. Thus any $ 4$ points of $ A_j$, no $ 3$ of them collinear, form a parallelogram.
Assume now there is a parallelogram in $ A_j$. Let it be $ ABCD$ and let $ P\equal{}AC\cap BD$. Suppose there exists some other point $ Q$ in$ A_j$. IF $ Q\notin AC$ and $ Q\notin BD$ then $ QP$ intersects some side of $ ABCD$, say $ AB$, in its interior - at $ R$. It follows $ R\in A_j$. But then $ P\in\textrm{Int}(RCD)$, impossible. Let now $ Q\in AC$, $ QA>QC$. Then $ QBDC$ isn't a parallelogram. Contradiction. Thus no such $ Q$ exists.
Then $ A_j$ is either the set $ \{A,B,C,D,P\}$ with $ ABCD$ parallelogram and $ P\equal{}AC\cap BD$, or no parallelogram and no quadrilateral in $ A_j$ exists. This means that $ A_j$ is a set of several collinear points or a set of collinear points plus one more point not collinear with the rest. In any of the possibilities for $ A_j$ it is quite clear that no subset of it can generate it as described in the statement of the problem. This means $ A_{j\minus{}1}\equal{}A_j$, or $ A_0\equal{}A_1\equal{}\ldots$, as desired.
Hope it's correct...


vector
geometry
3D geometry
sphere
parameterization
complex numbers
real analysis

Let $v$ be a vector field ( champ de vecteurs in french, I'm not sure I correctly translated the word) in $R^{n}$  such that whenever $x$ is on the unit sphere of $R^{n}$ one has $<v(x),x>\leq 0$. Suppose that a solution of $x'=v(x)$ starts its journey on the closed open ball. Prove that the poor solution is obliged to remain in this ball.

[quote="harazi"] one has $<v(x),x>\leq 0$. [/quote]
$\langle v(x), x \rangle$ looks nicer. :)

[quote="harazi"]Suppose that a solution of $x'=v(x)$ starts its journey on the [b]closed open ball[/b]. Prove that the poor solution is obliged to remain in this ball.[/quote]
Closed unit ball? Or something else?

My original thought was just to show that if we consider the distance from the origin, we see that we are always decreasing or 0 when it reaches 1, so it can't go more than 1. But what if it is a point of inflection (saddle point, whatever)? Seems that can't happen, but I am stuck. :(

Yes, you understood it correctly. Don't try to figure out what was in my head when I wrote closed open ball. It's probably because of the 16 oral exams I had to pass the last 2 weeks. Hopefully, it's done. This problem is difficult, I think.

It reminds me of [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=91736[/url] ;). By the way, you have to assume that the vector field is $C^{1}$ or, at least, Lipschitz. Otherwise the statement is just false.

Yes, it is C^1 class. And yes, there are strong similarities with that one!

This is going to be pretty long and clumsy, but I think it works.

Clearly, it suffices to consider trajectories which start at a point $x_{0}$ on the unit sphere such that $(v(x_{0}),x_{0})=0$, and to prove that we cannot have $x(0)=x_{0},\ \|x(t)\|>1$ for $t$ in some small interval $(0,\delta)$, or, equivalently, that if we do, $v$ cannot be $C^{1}$ (we'll prove that it cannot be locally Lipschitz around a point on the unit sphere actually).

Assume the opposite (for $x_{0}=(0,1,0,\ldots,0)$, say).

For our purposes, we can disregad the vector field $v$ outside some ball of radius $r>1$, i.e. we can assume that it vanishes outside a compact set. Such a $C^{1}$ vector field $v$ determines a unique one-parameter group of $C^{1}$ diffeomorphisms $\phi: \mathbb R\times\mathbb R^{n}\to\mathbb R^{n}$ which induces $v$ (i.e. for all $t\in\mathbb R,\ \phi_{t}: \mathbb R^{n}\to\mathbb R^{n}$ defined by $\phi_{t}(p)=\phi(t,p)$ is a diffeomorphism, $\phi_{t}\circ\phi_{s}=\phi_{t+s},\ \forall t,s\in\mathbb R$, and $\phi_{p}(t)=\phi(t,p)$ is a solution to $x'=v(x),x(0)=p$ for every $p\in\mathbb R^{n}$).

Now let $f$ be a ($C^\infty$, or $C^{1}$, or whatever) diffeomorphism of $\mathbb R^{n}$ onto itself which fixes the unit ball, and such that $f(\phi(t,x_{0}))=(1,0,\ldots,0)$ for $t\in(0,\delta)$. This is possible because for such small $t$ the trajectory lies outside the unit ball, by assumption. $\widetilde\phi=f\circ\phi\circ(\mbox{id}\times f)^{-1}: \mathbb R\times\mathbb R^{n}\to\mathbb R^{n}$ is easily seen to be a one-parameter group of diffeomorphisms, which induces some vector field $\widetilde v$.

Now consider the plane $\pi$ through the origin, $(1,0,\ldots,0)$, and $(0,1,\ldots,0)$. For each $x\in\pi$, replace $\widetilde v(x)$ with its projection onto $\pi$. The problem is now reduced to the plane (the vector field will be denoted by $v$ again, from now on), with the extra assumption that for small enough non-negative $u\in\mathbb R$ we have $v(u,1)=(1,0)$ (note that projection onto $\pi$ has not changed the fact that the vector field $v$ points inward on the boundary of the unit ball, which has now become the unit circle).

It suffices to show that no matter how small $\varepsilon>0$ is, there is no constant $L>0$ such that $\|v(x,\sqrt{1-x^{2}})-v(u,1)\|\le L\|(x,\sqrt{1-x^{2}})-(u,1)\|\ (*),\ \forall x,u\in[0,\varepsilon)$. It's easy to see that the LHS of $(*)$ is minimized if we choose $v(x,\sqrt{1-x^{2}})=(1-x^{2},-x\sqrt{1-x^{2}})$ (a drawing should make this clear; this just says that we take the vector with origin $(x,y)\in\mathbb S^{1}$ to be the projection of the vector $(1,0)$ with the same origin on the tangent in the same point to the circle). Now proving what we want should be just a simple computation.

Actually, in an oral test, I believe you may just say: when the solution reaches the boundary, its velocity points to the inner side of the unit ball, so it would not get out of the ball.

Then the professor will let you pass :)

That's your opinion. Anyway, excellent solution, Grobber.

I am really sorry I just overlooked the difficulty when the $<v(x),x> = 0$.  I have an idea to try to shorten the proof.

Let us consider a group of vector fields $v_\epsilon(x)=v(x)+\epsilon*(-x)$
Clearly $v_\epsilon(x)$ is still a vector field of class $C^{1}$.  Now for any $v_\epsilon(x)$, its restriction on the unit sphere points strictly inwards, i.e. $<v_\epsilon(x),x><0$.  In this case the solution $y_\epsilon$ corresponding to $v_\epsilon(x)$ lies totally inside the closed ball.  Now applying the theorem which says that the solution of a system of odes $C^{1}$  depends on the parameter $\epsilon$, we conclude that the assertion is correct letting $\epsilon\rightarrow 0$.

the same version is $x'=f(x)$ assume that all real part of $D(f)(0)$ is negative then origin is sink.and another nice result is $x'=Ax$ and $x'=Bx$ if $A$ and $B$ have the same number of eignevalues with negative real part ,then the trajectories assigned to these ode is topologically conjugate.(dont use tough theorem like Hartman Grobman :) )


algebra
polynomial
geometry
geometric transformation
reflection
real analysis
real analysis unsolved

Find the lowest degree polynomial that satisfies the following constraints:

i) F(0)=0

ii)F(1)=0

iii)The maximum of F on the interval (0,1) occurs at point c

iv) F(x) is positive on the interval (0,1)

Question 2)
find the lowest degree polynomial such that satisfies the previous constraints as well as:

v) F(x) is in the interval [0,1] for all x in (0,1)

Well, the fifth constraint is completely meaningless- just scale down something that satisfies the first four.

The answer depends strongly on $ c$. When $ c\equal{}\frac12$, it's $ 2$. When $ c$ is between $ \frac13$ and $ \frac12$ (or the reflection), it's 3. Fourth-degree polynomials get you out to $ \frac14$, and so on.

$ ax(1\minus{}x)(x\plus{}r)^k$
$ a(x\plus{}r)^{k\minus{}1}(\minus{}(k\plus{}2)x^2\plus{}(k\plus{}1\minus{}2r)x\plus{}r)\equal{}0$
$ \minus{}(k\plus{}2)c^2\plus{}(k\plus{}1)c\equal{}2rc\minus{}r$
$ r\equal{}\frac{c(k\plus{}1\minus{}(k\plus{}2)c)}{2c\minus{}1}$
We need $ r\ge 0$, so this gives a maximum at $ c$ as long as $ \frac12<c\le\frac{k\plus{}1}{k\plus{}2}$.
There's a similar construction on the other side.

It is not that simple, jmerry. Consider $ f(x) \equal{} x(1 \minus{} x)(x \minus{} 0.5)^2$. Its maxima are around $ 0.15$ and $ 0.85$, which beats $ 1/4$ or even $ 1/5$ for your 4th and 5th degree polynomials ;) I'm inclined to believe that the key word for this problem starts with Ch (or Tsch, whichever spelling you prefer)... :)