algebra
polynomial
calculus
derivative
superior algebra
superior algebra unsolved

If $ K \subset L$ of p degree, p prime than $ K \subset L$ is normal or separable??

$ K\equal{}\mathbb Q , L\equal{}K[\sqrt[3]2]$
$ K\equal{}\mathbb F_3(x) , L \equal{} K[\sqrt[3]x]$

isn't $ Q \subset Q[\sqrt[3]{2}]$ separable?

Read that as one counterexample for normal, and another for separable.

Maybe he's looking for a counterexample for both simultaneously.

Awful title, by the way.

yes for both at the same time... sorry for the title

Two theorems:
Every extension of a field of characteristic zero is separable.
Every finite extension of a finite field is normal, since the field with $ p^n$ elements is the splitting field of $ x^{p^n}\minus{}x$ over $ \mathbb{Z}/p\mathbb{Z}$.
Actually, I can't think of a single finite non-normal extension of a field of finite characteristic.

[quote="jmerry"]Actually, I can't think of a single finite non-normal extension of a field of finite characteristic.[/quote]
$ K\equal{}\mathbb F_5(x) , L \equal{} K(\sqrt[3]x)$?

It's still separable, of course.

$ K\equal{}\mathbb F_3(x), L \equal{} K[\sqrt[15]x]$ works for both.

There does not exist a field extension of prime degree that is neither separable nor normal.

Proof: Let $ \mathbb{F} \subset \mathbb{K}$ be an inseparable extension of prime degree.  Since finite extensions are algebraic, we have $ \mathbb{K} \equal{} \mathbb{F}(\alpha)$ for some $ \alpha\in K$.  Let $ f(x)$ be the minimal polynomial for $ \alpha$ over $ \mathbb{F}$, and let $ p > 0$ be the characteristic of $ \mathbb{F}$.

Since the derivative of an inseparable irreducible polynomial is $ 0$, and $ f$ has prime degree, we must have $ f(x) \equal{} x^p \minus{} c$ for some $ c\in F$.  But then $ \alpha^p \equal{} c$ and $ x^p \minus{} c \equal{} (x \minus{} \alpha)^p$ splits in $ \mathbb{K}$, so $ \mathbb{K}$ is normal over $ \mathbb{F}$.


this ones a classic
4 bugs are situated in the corners of a 1 by 1 square
each bug moves towards the one clockwise of it a at a rate of 1 unit per minute
how long will it take for the bugs to reach eachother?

[quote="xxreddevilzxx"][hide="hint"]
think of the path... what kind of shape does it create??[/hide][/quote]Do the bugs make a spiral?

yea... eryaman, you are on track.

To Scrambled: You should have said, "how long does it take for 4 bugs to meet together", instead of saying "when will the bugs meet at the center". That gave away too mcuh.... well that and my hint ;)


rotation
combinatorics unsolved
combinatorics

A pawn is placed on a square of a $ n \times n$ board. There are two types of legal
moves: (a) the pawn can be moved to a neighbouring square, which shares a common side with the current square; or (b) the pawn can be moved to a  neighbouring square, which shares a common vertex, but not a common side with the current square. Any two consecutive moves must be of different type. Find all integers $ n \ge 2$, for which it is possible to choose an initial square and a sequence of moves such that the pawn visits each square exactly once (it is not required that the pawn returns to the initial square).

For odd $ n = 2k + 1$, color the board black and white in a checkerboard fashion, and let the squares be named $ (i,j)$, where $ 1 \le i \le n$ denotes the row and $ 1 \le j \le n$ denotes the column. Then move (a) causes the pawn to move to a square of an alternate color, while move (b) causes the pawn to move to a square of the same color. Note that from any square of a color, either the move before or after (but not both) must be to another adjacent square of the same color.

Suppose a sequence exists where the starting and ending points are not opposite corners; then we rotate the board such that the starting point and ending points are not on the first row. WLOG let the corners be all black. The only black square next to corner $ (1,1)$ is $ (2,2)$, and similarly to $ (1,2k + 1)$ is $ (2,2k)$. Then, in our sequence, we must have somewhere
\begin{align*} \mathrm{white} - (1,1) & \leftrightarrow (2,2) - \mathrm{white} \\
\mathrm{white} - (1,2k + 1) & \leftrightarrow (2,2k) - \mathrm{white} \tag{*} \end{align*}
Since we have already traversed through $ (2,2)$, then the only available black square adjacent to $ (1,3)$ is $ (2,4)$. Repeating this argument, the only black square adjacent to black square $ (1,2m + 1)$ is $ (2,2m + 2)$, and so the only black square adjcent to $ (1,2k - 1)$ is $ (2,2k)$, contradicting $ (*)$.

If a sequence exists where the starting and ending points are opposite corners, WLOG let the starting point be $ (1,1)$, and use a similar contradiction based upon the black squares in the first row and first column.

For $ \boxed{\mathrm{even}\ n}$, we can repeat the following sequence:
[asy]int row = 8, col = 2;
for(int i = 0; i < row + 1; ++i)
 draw((2i,0)--(2i,2*col));
for(int j = 0; j < col + 1; ++j)
 draw((0,2j)--(2*row,2j));

for(int i = 1; i < row/2; ++i){
 draw((4i+1,1)--(4i+3,3));
 draw((4i+1,3)--(4i+3,1));
}
for(int i = 1; i < row/2; ++i){
 draw((4i-1,1)--(4i+1,1));
 draw((4i-1,3)--(4i+1,3));
}
dot((1,1)); draw((1,1)--(1,3)--(3,1)); draw((2*row-1,1)--(2*row-1,3));[/asy]


limit
calculus
logarithms
calculus computations

Same as the sequence limit $ \lim_{n\to\infty}n^{1/n},$

 -- and, as I told my calculus class when we were studying the convergence of series, "If you're going to be using the root test at all, you need to just [i]know[/i] this one."

Here is the answer :) ::

lim(x -> infinity) { x^ 1/x} = lim  e ^ { log (x^1/x) to the base e}
                                     = lim  e^{ (1/x)*log x to the base e}
                                     =1.

The base $ e$ logarithm is often denoted by $ \ln$- and when it isn't, it's just called $ \log$. Shorter notation is a good thing.

I know about ln.
I didn't use that may be someone don't know.

anyways thanks.

hello, using L'Hospital's rule we have ${ \lim_{x \to \infty}x^{\frac{1}{x}}=e^{\lim_{x \to \infty}\frac{\ln(x)}{x}}}=e^{\lim_{x \to \infty}\frac{1}{x}}=e^0=1$.
Sonnhard.


Two integers are randomly selected from the set of integers greater than or equal to $ \minus{}5$ and less than or equal to 5. The two numbers need not be different. What is the probability that the sum of the two integers is less than their product?

Case 1: Both are negative
All work [b](25)[/b]
Case 2: 1 positive 1 negative
The sum is going to be greater than the product:(
Case 2: Both positive
All work except for 1 1, 1 2, 1 4, 1 5, 2 1, 3 1, 4 1, 5 1, and 2 2 [b](16)[/b]
Case 3:
0 and Negative
All work [b](5)[/b]
Case 4:
0 and Positive
Don't work
Case 5:
0 and 0, don't work
$ 25\plus{}16\plus{}5\equal{}46$ favorable.
$ 11^2$ total possibilites.
$ \boxed{\frac{46}{121}}$ is answer

Um, AoD, you must have made some type of error, because apparently the answer is 50/121.... o.O

Probably an easier way but...
[hide="Solution"]Let the two numbers we choose be $ a$ and $ b$ (it may be that $ a \equal{} b$).  We want to find all ordered pairs $ (a,b)$, where $ a$ and $ b$ are integers between $ \minus{} 5$ and $ 5$, inclusive, such that
\[ ab > a \plus{} b\]
We can manipulate the inequality so that we have
\[ ab \minus{} a \minus{} b \plus{} 1 > 1\]
Which we can factor into
\[ (a \minus{} 1)(b \minus{} 1) > 1\]
This happens when
(1)  $ a \minus{} 1$ and $ b \minus{} 1$ are both positive (and $ a\ne 2$ and $ b\ne 2$) or
(2)  $ a \minus{} 1$ and $ b \minus{} 1$ are both negative (and $ a\ne 0$ and $ b\ne 0$).

[hide="For case (1)..."] we have a total of $ 4$ choices for $ a$ ($ 2$ to $ 4$ inclusive) and $ 4$ choices for $ b$ ($ 2$ to $ 4$, inclusive).  So it appears there are $ 4\cdot 4$ different ordered pairs $ (a,b)$.  However, we cannot count the ordered pair $ (2,2)$, so we only have $ 16 \minus{} 1 \equal{} 15$ ordered pairs for case $ (1)$.[/hide]

[hide="For case (2)..."] we have a total of $ 6$ choices for $ a$ ($ \minus{} 5$ to $ 0$, inclusive) and $ 6$ choices for $ b$ ($ \minus{} 5$ to $ to$, inclusive).  So it appears there are $ 6\cdot 6$ different ordered pairs $ (a,b)$.  However, we cannot count the ordered pair $ (0,0)$, so we only have $ 36 \minus{} 1 \equal{} 35$ ordered pairs for case $ (2)$.[/hide]

Since there are a total of $ 11^2 \equal{} 121$ ordered pairs, the probability that $ ab > a \plus{} b$ is thus $ \frac {15 \plus{} 35}{121} \equal{} \boxed{\frac {50}{121}}$[/hide]

Much simpler than all above.

Solution: Note that the probability $a+b>ab$ is same as probability $a+b<ab$ by smmetry. So, we only need to find the case where $a+b=ab$, which occurs only for $a=b=0$ in our case. Thus the probability is $\frac{1-\frac{1}{121}}{2}=\frac{50}{121}$.


The sides $ [AB]$ and $ [AC]$ of the triangle $ ABC$ are tangent to the incircle with center $ I$ of the $ \triangle ABC$ at the points $ M$ and $ N$, respectively. The internal bisectors of the $ \triangle ABC$ drawn form $ B$ and $ C$ intersect the line $ MN$ at the points $ P$ and $ Q$, respectively. Suppose that $ F$ is the intersection point of the lines $ CP$ and $ BQ$. Prove that $ FI\perp BC$.

[hide]
It suffices to show that $ \angle BPC \equal{} \angle BQC \equal{} 90$. Then, if $ H$ is the intersection of $ BP$ and $ CQ$, it would follow that $ F$ is the orthocenter of $ \triangle BHC$, from which the result follows.

WLOG, we will show that $ \angle BQC\equal{}90$; then, $ \angle BPC\equal{}90$ follows by symmetry.

Let $ B'$ be on $ AC$ such that $ CB \equal{} CB'$. Let $ BB'$ intersect $ MN$ at $ Q'$; I claim that $ Q'$ is the midpoint of $ BB'$. But that would mean $ CQ'$ is the angle bisector of $ \angle ACB$, so then $ Q'\equal{}Q$ and $ \angle BQC \equal{} 90$, whence we'd be done.

To show the claim, apply Menelaus' theorem to $ \triangle ABB'$ and line $ MQN$. Then,

\[ \frac{AM}{MB} \cdot \frac{BQ}{QB'} \cdot \frac{NB'}{NA} \equal{} 1\]
\[ \Longleftrightarrow \frac{ \frac{b\plus{}c\minus{}a}{2} }{ \frac{c\plus{}a\minus{}b}{2} } \cdot \frac{BQ}{QB'} \cdot \frac{ a\minus{}\frac{a\plus{}b\minus{}c}{2} }{ \frac{b\plus{}c\minus{}a}{2} } \equal{} 1\]
\[ \Longleftrightarrow \frac{ b\plus{}c\minus{}a }{ c\plus{}a\minus{}b } \cdot \frac{BQ}{QB'} \cdot \frac{ c\plus{}a\minus{}b }{ b\plus{}c\minus{}a } \equal{} 1\]

So, $ BQ \equal{} QB'$, as desired.

[/hide]

It is very simple to show that  $ \angle BPC =\angle BQC = 90 $ .    Notice that   $ \angle ANM =\angle AMN = 90 -1/2 \angle A =\angle PNC. $
Let $ \{L\}= BP\cap AC $ , then $ \angle BLC = 180 - (\angle B + 1/2\angle C ) $, but $ \angle LIC = 180 - \angle BLC - \angle C/2 = \angle B/2 + \angle C/2 = 90 - \angle A/2 = \angle PNC $, it follows that $ INPC $ is cyclic, so $ \angle BPC = \angle IPC = \angle INC = 90 $. Similarly we show that $\angle BQC = 90$. From these results we claim that $I$ is the orthocenter of the $\triangle BFC$ ,where $I$ is the incenter of $\triangle ABC$.

[color=darkred][b][u]PP[/u].[/b] The sides $ [AB]$ and $ [AC]$ of the triangle $ ABC$ are tangent to the incircle $w=C(I)$ of the $ \triangle ABC$ at the points $ M$ and $ N$, respectively. The internal bisectors of the $ \triangle ABC$ drawn form $ B$ and $ C$ intersect the line $ MN$ at the points $ P$ and $ Q$, respectively. Suppose that $ F$ is the intersection point of the lines $ CP$ and $ BQ$. Prove that $ FI\perp BC$.[/color]

[b][u]Proof[/u].[/b] Suppose w.l.o.g $N\in MP$ and $Q\in (MN)$ . Therefore,

$\blacktriangleright\ \left\{\begin{array}{ccc}
m\left(\widehat{PIC}\right)=180^{\circ}-m\left(\widehat{BIC}\right)=180^{\circ}-\left(90^{\circ}+\frac A2\right)=90^{\circ}-\frac A2 & \implies & m\left(\widehat{PIC}\right)=90^{\circ}-\frac A2\\\\
m\left(\widehat{PNC}\right)=m\left(\widehat{ANM}\right)=90^{\circ}-m\left(\widehat{NAI}\right)=90^{\circ}-\frac A2 & \implies & m\left(\widehat{PNC}\right)=90^{\circ}-\frac A2\end{array}\right|\implies$

$PNIC$ is cyclically $\iff$ $m\left(\widehat {IPC}\right)=m\left(\widehat{INC}\right)=90^{\circ}\iff$ $PC\perp PI\iff \boxed{PC\perp PB}\ (1)$ .

$\blacktriangleright\ \left\{\begin{array}{ccc}
m\left(\widehat{QIB}\right)=180^{\circ}-m\left(\widehat{BIC}\right)=180^{\circ}-\left(90^{\circ}+\frac A2\right)=90^{\circ}-\frac A2 & \implies & m\left(\widehat{QIB}\right)=90^{\circ}-\frac A2\\\\
m\left(\widehat{AMN}\right)=90^{\circ}-m\left(\widehat{MAI}\right)=90^{\circ}-\frac A2 & \implies & m\left(\widehat{AMN}\right)=90^{\circ}-\frac A2\end{array}\right|\implies$

$QMBI$ is cyclically $\iff$ $m\left(\widehat {IQB}\right)=m\left(\widehat{IMB}\right)=90^{\circ}\iff$ $QB\perp QI\iff \boxed{QB\perp QC}\ (2)$ .

In conclusion, from the relations $(1)$ and $(2)$ obtain that the incenter $I$ is the orthocenter of $\triangle BFC$ , i.e. $FI\perp BC$ .

[b][u]Remark[/u].[/b] If $D\in BC\cap w$ , then $\triangle DQP\sim\triangle ABC$ and $I$ is incenter of $\triangle DQP$ .

Dear Mathlinkers,
you can see
http://jl.ayme.pagesperso-orange.fr/   vol. 4, p. 5

Sincerely
Jean-Louis


AMC
USA(J)MO
USAMO
geometry
circumcircle
geometry proposed

Take triangle $ ABC$ with circumcenter $ O$.  Denote the midpoints of $ BC$, $ AC$, and $ AB$ to be $ A_1$, $ B_1$, and $ C_1$ respectively.  Take $ A_2$ on $ OA_1$ such that $ \triangle OAA_1\sim \triangle OA_2A$.  Define $ B_2$ and $ C_2$ similarly.  Denote the intersections between $ A_2B_2$ and $ BC$ to be $ A_3$, between $ A_2B_2$ and $ AC$ to be $ B_3$, between $ A_2C_2$ and $ BC$ to be $ A_4$, between $ A_2C_2$ and $ AB$ to be $ C_4$, between $ C_2B_2$ and $ AB$ to be $ C_3$, and between $ C_2B_2$ and $ AC$ to be $ B_4$.  Show that $ (AB_4)(A_3C)(BC_4)\equal{}(CB_3)(BA_4)(AC_3)$.

Nobody interested?

Sorry for my notation.  Allow me to rephrase the problem (kind of):
Take triangle $ ABC$ with circumcenter $ O$.  Denote the midpoints of $ BC$, $ AC$, and $ AB$ to be $ A_1$, $ B_1$, and $ C_1$ respectively.  Take $ A_2$ on $ OA_1$ such that $ \triangle OAA_1\sim \triangle OA_2A$.  Define $ B_2$ and $ C_2$ similarly.  Denote the following points: $ A_2B_2\cap BC \equal{} A_3$, $ A_2B_2\cap AC \equal{} B_3$, $ A_2C_2\cap BC \equal{} A_4$, $ A_2C_2\cap AB \equal{} C_4$, $ B_2C_2\cap AB \equal{} C_3$, and $ B_2C_2\cap AC \equal{} B_4$.  Prove that
\[ (AB_4)(A_3C)(BC_4) \equal{} (CB_3)(BA_4)(AC_3)
\]

I hope this makes the problem clearer and hopefully less messy  :oops:

Hmm....problem is weird;;; :|
From $ \triangle OAA_1 \sim \triangle OA_2A$  we can know that $ OA^2\equal{}OA_1 \times OA_2$....
and also from $ OA\equal{}OC$, $ OC^2\equal{}OA_1 \times OA_2$
so we can say $ \triangle OA_1C \sim \triangle OCA_2$ and from this, $ \angle OA_1C\equal{}\angle OCA_2\equal{}\dfrac{\pi}{2}$

we can do above process symmetrically....then $ A_2$ and $ C$ and $ B_2$ are collinear because $ \angle OCA_2\equal{}\angle OCB_2\equal{}\dfrac{\pi}{2}$......
then,,,,,$ A_3\equal{}C\equal{}B_3$ and $ (AB_4)(A_3C)(BC_4)\equal{}(CB_3)(BA_4)(AC_3)\equal{}0$
it seems strange but... anyway;;;; the result is right.... :maybe:

[quote="limsk1"]Hmm....problem is weird;;; :|
From $ \triangle OAA_1 \sim \triangle OA_2A$  we can know that $ OA^2 \equal{} OA_1 \times OA_2$....
and also from $ OA \equal{} OC$, $ OC^2 \equal{} OA_1 \times OA_2$
so we can say $ \triangle OA_1C \sim \triangle OCA_2$ and from this, $ \angle OA_1C \equal{} \angle OCA_2 \equal{} \dfrac{\pi}{2}$

we can do above process symmetrically....then $ A_2$ and $ C$ and $ B_2$ are collinear because $ \angle OCA_2 \equal{} \angle OCB_2 \equal{} \dfrac{\pi}{2}$......
then,,,,,$ A_3 \equal{} C \equal{} B_3$ and $ (AB_4)(A_3C)(BC_4) \equal{} (CB_3)(BA_4)(AC_3) \equal{} 0$
it seems strange but... anyway;;;; the result is right.... :maybe:[/quote]
Ah... I never realized that they were collinear  :blush:
Here was my solution:
[hide="Solution"]
We notice that since $ \triangle OAA_2\sim \triangle OA_1A\implies \frac {OA}{OA_2} \equal{} \frac {OA_1}{OA}\implies OA^2 \equal{} OA_1\times OA_2$.  Similarly, $ OB^2 \equal{} OB_1\times OB_2$.  Since $ OA \equal{} OB$, we have that $ OA_1\times OA_2 \equal{} OB_1\times OB_2$, so $ A_2A_1B_1B_2$ is cyclic.  Denote the intersections between $ A_2B_1$ and $ A_1A_3$ to be $ X$ and the intersection between $ B_1B_3$ and $ A_1B_2$ to be $ Y$.  Now, this means that $ 90 \plus{} \angle YA_1X \equal{} \angle A_2A_1B_2 \equal{} \angle A_2B_1B_2 \equal{} 90 \plus{} \angle XB_1Y$, so we obtain cyclic quaderilateral $ XYB_1A_1$.  Thus, $ \angle A_1YX \equal{} \angle XB_1A_1 \equal{} \angle A_2B_1A_2 \equal{} \angle A_2B_2A_1\implies XY\parallel A_2B_2$.  Now, since $ A_1$ and $ B_1$ are the midpoints of $ BC$ and $ AC$ respectively, we have that $ A_1B_1\parallel AB$.  Thus, $ \angle ABC \equal{} \angle B_1A_1X \equal{} 180 \minus{} \angle B_1YX \equal{} 180 \minus{} \angle A_3B_3B_1$.  Thus, $ A_3B_3AB$ is cyclic.  Similarly, $ BCB_3C_3$ and $ ACA_3C_3$ are cyclic.  Thus, $ \angle BAC \equal{} \angle B_3A_3C \equal{} \angle A_4A_3A_2$ and $ \angle BAC \equal{} \angle BA_4C_4 \equal{} \angle A_2A_4A_3$.  So, $ \angle A_2A_4A_3 \equal{} \angle A_4A_3A_2$, and thus $ \triangle A_4A_3A_2$ is isoceles.  Now, since $ A_1$ is the midpoint of $ BC$, we have that $ A_2A_1\perp BC\implies A_4A_1 \equal{} A_3A_1\implies BA_4 \equal{} CA_3$.  Similarly, $ CB_3 \equal{} AB_4$ and $ AC_3 \equal{} BC_4$.  From this we can see that $ (AB_4)(A_3C)(BC_4) \equal{} (CB_3)(BA_4)(AC_3)$.  Now, noticing that $ A_2CB_2$ is a line is a much faster solution... I should have constructed the figure...  :( [/hide]


[quote="FrankieBlueEyes"]all seven digit telephone numbers in a town begin with 245 how many telephone numbers may be assigned in the town if the last for digits do not begin or end with a zero?
HelP!! please[/quote]

This is just a normal Combinatorics question.

Hope I don't make stupid mistakes

[hide]There are $9$ choices for the the fouth and the last digits, and $10$ choices for the other two. So there are totally $9*10*10*9=8100$ numbers assigned in the town.[/hide]

hint:

how many numbers can the first number be, then the second, third?...


inequalities
function
inequalities proposed

Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a\plus{}b\plus{}c)^6>64abc(a\plus{}b)(b\plus{}c)(c\plus{}a)$.

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
Stronger ?
http://www.mathlinks.ro/viewtopic.php?t=216979

$ a\plus{}(b\plus{}c) \ge 2\sqrt{a(b\plus{}c)}$
$ b\plus{}(c\plus{}a) \ge 2\sqrt{b(c\plus{}a)}$
$ c\plus{}(a\plus{}b) \ge 2\sqrt{c(a\plus{}b)}$

So, $ (a\plus{}b\plus{}c)^6 > 2^6 \left(\sqrt{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}\right)^2 \equal{} 64abc(a\plus{}b)(b\plus{}c)(c\plus{}a)$

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
It is really easy,kunny
Apply AMGM $ abc(a \plus{} b)(b \plus{} c)(c \plus{} a)\le \frac {8}{3^6}(a \plus{} b \plus{} c)^6$and the best value is $ \frac {3^6}{8}$
The equal holds when $ a \equal{} b \equal{} c$

What about the ineq of Honey_S? It is not solvable in the same procedure....is it?

[quote="Sunkern_sunflora"]It is not solvable in the same procedure....is it?[/quote]
I suppose not...

I was thinking if we can try mixing variables method? Mixing $ b,c$ to obtain $ \sqrt{bc}$? I might be wrong....

[quote="Sunkern_sunflora"]I was thinking if we can try mixing variables method?[/quote]
Yes it could maybe. But I thought you were talking about the easy AM-GM solutions of the other ineq.
But if you are going to use MV on $ \sqrt {bc}$ you'll have to use it on $ (b^2 \plus{} c^2)$ too along with $ (b \plus{} c)$, but both decreases. Which might make it a bit complicated and possibly not always true. However I've not tried on this.

Yes....I have not thought of that....anyway, I am still thinking....there ought to be a good substitution.
 :)

P/s: Tomar problem ta dekhechi.....daarooon jinish ber korecho. Kintu tomar mathae elo kikore? Etto sokto problem....amar dara hobe bole mone hoy na..... tobu vabbo nischoi. Thank u!

[hide="Off topic"][quote="Sunkern_sunflora"]P/s: Tomar problem ta dekhechi.....daarooon jinish ber korecho. Kintu tomar mathae elo kikore? Etto sokto problem....[/quote]Khub akta kothin kono theorem lage na. Tobe khub i unconventional. Common techniq ba orocom kichu die hobe na. Anik dada r pochondo hobe, NT Genius. Ami ektu primes distribution and related stuff nie kichu kichu study korchilam, tokhon eta berieche, ar ekta problem shape korche, tobe ota akhon o parini. Purota figure out korle otao post korbo khon.[/hide]

[quote="Honey_S"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
\[ \left( {a \plus{} b \plus{} c} \right)^5 \ge 81abc\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)
\]
[/quote]
It is equivalent to
\[ (a\plus{}b\plus{}c)^6 \ge 81abc(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\]
Since $ 3abc(a\plus{}b\plus{}c) \le (ab\plus{}bc\plus{}ca)^2$, it suffices to prove that
\[ (a\plus{}b\plus{}c)^6 \ge 27(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)^2\]
By AM-Gm Inequality, we have
\[ (a\plus{}b\plus{}c)^6 \equal{}[(a^2\plus{}b^2\plus{}c^2)\plus{}(ab\plus{}bc\plus{}ca)\plus{}(ab\plus{}bc\plus{}ca)]^3 \ge 27\cdot (a^2\plus{}b^2\plus{}c^2)\cdot (ab\plus{}bc\plus{}ca) \cdot (ab\plus{}bc\plus{}ca) \equal{}27(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)^2\]
We have done. :)

[quote="Honey_S"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
\[ \left( {a \plus{} b \plus{} c} \right)^5 \ge 81abc\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)
\]
[/quote]
More easy,Fix $ a \plus{} b \plus{} c,\sum ab$ here is a decrease function with variables $ abc$Use ABC method,we only need check the cases $ a \equal{} b$or $ c \equal{} 0$.It is obivious

[quote="Allnames"][quote="Sunkern_sunflora"][quote="Allnames"]$ \sum ab$ here is a [b]crease[/b] function with variables $ abc$[/quote]
Do you want to mean increase or decrease?? :maybe:[/quote]
decrease,Thanks
To sorry,I will post some stronger problem than Honey _suck' one
$ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^3 \plus{} 48abc(ab \plus{} bc \plus{} ca)\ge 25abc(a \plus{} b \plus{} c)^2$[/quote]

Firstly, the inequality that I have posted $ \left( {a \plus{} b \plus{} c} \right)^5  \ge 81abc\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)$ is not mine, it is an inequality of nguoivn and Vasc.  :oops:
And the next, the inequality that you have posted, assume $ a\plus{}b\plus{}c\equal{}1$ then it becomes
\[ ab \plus{} bc \plus{} ca \plus{} 48abc\left( {ab \plus{} bc \plus{} ca} \right) \ge 25abc
\]
which can be proved by the theorem $ r \le \frac{{\left( {1 \minus{} q} \right)^2 \left( {1 \plus{} 2q} \right)}}{{27}}$.  :)

[quote="Allnames"]
To sorry,I will post some stronger problem than Honey _suck' one
$ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^3 \plus{} 48abc(ab \plus{} bc \plus{} ca)\ge 25abc(a \plus{} b \plus{} c)^2$[/quote]
It's equivalent to $ \sum_{cyc}c(a\minus{}b)^2(a\plus{}b\minus{}3c)^2\geq0.$

[quote="arqady"][quote="Allnames"]
To sorry,I will post some stronger problem than Honey _suck' one
$ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^3 \plus{} 48abc(ab \plus{} bc \plus{} ca)\ge 25abc(a \plus{} b \plus{} c)^2$[/quote]
It's equivalent to $ \sum_{cyc}c(a \minus{} b)^2(a \plus{} b \minus{} 3c)^2\geq0.$[/quote]
Thank for your nice proof,My solution is similar to honeysuck'one,and of course,the problem is not mine

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
Stronger ? :)
$ (a \plus{} b \plus{} c)^6 \ge\ 32(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)$

[quote="onlylove_math"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
Stronger ? :)
$ (a \plus{} b \plus{} c)^6 \ge\ 32(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)$[/quote]

Your inequality is true.  :)

But it is not stronger than $ \left( {a \plus{} b \plus{} c} \right)^6  \ge 64abc\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)$.

For example with the inequality $ \left( {a^2  \plus{} b^2 } \right)\left( {b^2  \plus{} c^2 } \right)\left( {c^2  \plus{} a^2 } \right) \ge 2abc\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)$, try $ a\equal{}b\equal{}c\equal{}1$.  :)

[quote="Honey_S"][quote="onlylove_math"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
Stronger ? :)
$ (a \plus{} b \plus{} c)^6 \ge\ 32(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)$[/quote]

Your inequality is true.  :)

But it is not stronger than $ \left( {a \plus{} b \plus{} c} \right)^6 \ge 64abc\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)$.

For example with the inequality $ \left( {a^2 \plus{} b^2 } \right)\left( {b^2 \plus{} c^2 } \right)\left( {c^2 \plus{} a^2 } \right) \ge 2abc\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)$, try $ a \equal{} b \equal{} c \equal{} 1$.  :)[/quote]
Yes , But it isn't also looser than kunny 's one  :)
 It 's easy and I have known a proof by SS  :)

[quote="onlylove_math"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
Stronger ? :)
$ (a \plus{} b \plus{} c)^6 \ge\ 32(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)$[/quote]
No! $ 32(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)\geq 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$ is wrong.
What about the following inequality?
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a \plus{} b \plus{} c)^8\geq64(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} c^2)
\]

[quote="arqady"][quote="onlylove_math"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers.

Prove that $ (a \plus{} b \plus{} c)^6 > 64abc(a \plus{} b)(b \plus{} c)(c \plus{} a)$.[/quote]
Stronger ? :)
$ (a \plus{} b \plus{} c)^6 \ge\ 32(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)$[/quote]
No! Try $ a \equal{} b \equal{} c \equal{} 1.$
But the following inequality more stronger then your.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a \plus{} b \plus{} c)^8\geq64(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} c^2)
\]
[/quote]
Also see here for the first solution,
http://www.mathlinks.ro/viewtopic.php?p=1151957#1151957
And there also a proof with the AM-GM inequality. :)
But how it can be stronger than the onlylove_math's one since $ \frac {{\left( {a \plus{} b \plus{} c} \right)^8 }}{{2\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}} \le \left( {a \plus{} b \plus{} c} \right)^6$ is not true.

[quote="arqady"][quote="Honey_S"]
But how it can be stronger than the onlylove_math's one since $ \frac {{\left( {a \plus{} b \plus{} c} \right)^8 }}{{2\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}} \le \left( {a \plus{} b \plus{} c} \right)^6$ is not true.[/quote]
I have checked  it already.[/quote]

$ \frac {{\left( {a \plus{} b \plus{} c} \right)^8 }}{{2\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}} \le \left( {a \plus{} b \plus{} c} \right)^6 \Leftrightarrow a^2 \plus{} b^2 \plus{} c^2 \ge 2ab \plus{} 2bc \plus{} 2ca$ . How can it be true, arqady.  :?:

Honey_S, I have fixed it already.
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1245090#1245090


The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
\[
\begin{array}{cccc}
& 6 & 4 & 1 \\
& 8 & 5 & 2 \\
+ & 9 & 7 & 3 \\ \hline
2 & 4 & 5 & 6
\end{array}
\]

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

[quote="Silverfalcon"]The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?

\[ \tabular{cccc} & 6 & 4 & 1 & \\ & 8 & 5 & 2 & \\ + & 9 & 7 & 3 & \hline \\ 2 & 4 & 5 & 6 \end{tabular}  \]

A. 4
B. 5
C. 6
D. 7
E. 8[/quote]

[hide]
Eh.. D?  You have to change the second column, and the largest one there is 7.

And I don't see how changing 8 helps..
[/hide]

[quote="ccy"]And I don't see how changing 8 helps..
[/quote]

Who told you that it helped?  :?

[hide]
I'd say D, since 4+5+7 does not equal 15, it equals 16, so you'd have to make the 7 a 6.[/hide]

d.
you have 2 change the second column and 7 is the biggest number in it


induction
number theory unsolved
number theory

assume that X is set of the integer which they are in this form:
a(2k)10^2k+a(2k-2)*10^(2k-2)+....+a2*10^2+a0
k is nonnegative integer a(2i) are in {1,2,...,9}. prove that each number in form 2^p*3^q(p and q are nonnegative integers)divides atleat one of the member of of X

First, we prove by induction that we may select the first 2n-1 digits of the number such that it is divisible by 2^(2n).
For n=1, this is trivially true, since 4 is divisible by 4.

Assume that we can find such a sequence of initial digits for n=k.  Then by hypothesis, the number is of the form a*2^2k for some integer a.  We increase it to length k+1 by adding b*10^2k where b is some digit.  a*2^2k+b*10^2k=2^2k(a+b*5^k).  But a+b^5k==a+b (mod 4).  And since there are digits in every residue class mod 4, we may select b such that a+b==0 (mod 4), thus completing the induction.

So, we may find a block of digits B divisible by 2^p for any p.  But then the number (B)(0)(B)(0)(B) (juxtaposing (B) and 0 together)=B(1+10^(k+1)+10^(2k+2)) (for some k, depending on the length of B)
==B(1 + 1 + 1)==0 (mod 3).
Suppose we can find some block of digits X divisible by 3^k.  Then by the exact same procedure as above, (X)(0)(X)(0)(X) is divisible by 3^(k+1), proving by induction that we may find some block of digits of the required type divisible by 3^q for any q.

Combining the results of these two inductions, since (3^q, 2^p)=1, we may find some number of the required type divisible by 3^q 2^p for any p and q in N.


inequalities
algebra
polynomial
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{}\frac{a^2b\plus{}b^2c\plus{}c^2a}{ab^2\plus{}bc^2\plus{}ca^2} \ge \frac{5}{2}\]

[quote="mitdac123"]Try
\[ c \equal{} 0
\]
:P[/quote]
What do you mean?
When $ c\equal{}0$, the inequality becomes
\[ \frac{a}{b}\plus{}\frac{b}{a}\plus{}\frac{a^2b}{ab^2} \ge \frac{5}{2}\]
\[ \Leftrightarrow \frac{2a}{b} \plus{}\frac{b}{a} \ge \frac{5}{2}\]
which is true since
\[ LHS \ge 2\sqrt{2} >\frac{5}{2}\]
;)

[quote="mitdac123"]:P.
* $ \sum a^2b \ge \sum ab^2$ , this ineq is obvious true by Nesbitt's ineq .
* $ \sum a^2b \le \sum ab^2$
+$ \rightarrow \frac {\sum a^2b }{\sum ab^2} - 1 \ge - \frac {3(a - b)(b - c)(c - a)}{2(a + b)(a + c)(b + c)}$
+$ \sum \frac {a}{b + c} - \frac {3}{2} = \frac {\sum (2a + b + c)(a - b)(a - c)}{2(a + b)(b + c)(c + a)}$
=>${ LHS \ge \frac {1}{(a + b)(b + c)(c + a)} [ \sum (a + b)(a - c)(a - b)} \ge 0$;)[/quote]
But the inequality
\[ \frac {\sum a^2b }{\sum ab^2} - 1 \ge - \frac {3(a - b)(b - c)(c - a)}{2(a + b)(a + c)(b + c)}
\]
is not true.
[quote="Allnames"]we can proved it by SS With $ c = min{a,b,c}$ and $ a + b + c = 1$
Use $ \sum \frac {a}{b + c} - \frac {3}{2} = (a - b)^2 \frac {1}{(a + c)(b + c)} + \frac {a + b + 2c}{2(a + b)(b + c)(c + a)}(a - b)(b - c)$
and $ \frac {a^2b + b^2c + c^2a}{ab^2 + bc^2 + ca^2} - 1 = \frac {(a - b)(b - c)(a - c)}{ab^2 + bc^2 + ca^2}$
the ineq is    $ M{a - b)^2 + N(a - c)(b - c)}$ there
$ M = \frac {1}{(a + c)(b + c)}),N = \frac {a + b + 2c}{2(a + b)(b + c)(c + a)} + \frac {(a - b)}{ab^2 + bc^2 + ca^2}$
It is very éay to find $ N\geq\ 0$ so the proof is finished[/quote]
The inequality $ N \ge 0$ is not true, try $ c=0,a=1,b=3$ ;)

Oh man . I have a mistake :) .If $ \frac {a}{b} \le 1 \rightarrow \frac {a \plus{} c}{b \plus{} c} \ge \frac {a}{b}$  , by I have thought  $ \frac {a}{b} \le 1 \rightarrow \frac {a \plus{} c}{b \plus{} c} \le \frac {a}{b}$  :(

[quote="onlylove_math"][quote="can_hang2007"]
[quote="Allnames"]we can proved it by SS With $ c = min{a,b,c}$ and $ a + b + c = 1$
Use $ \sum \frac {a}{b + c} - \frac {3}{2} = (a - b)^2 \frac {1}{(a + c)(b + c)} + \frac {a + b + 2c}{2(a + b)(b + c)(c + a)}(a - b)(b - c)$
and $ \frac {a^2b + b^2c + c^2a}{ab^2 + bc^2 + ca^2} - 1 = \frac {(a - b)(b - c)(a - c)}{ab^2 + bc^2 + ca^2}$
the ineq is    $ M{a - b)^2 + N(a - c)(b - c)}$ there
$ M = \frac {1}{(a + c)(b + c)}),N = \frac {a + b + 2c}{2(a + b)(b + c)(c + a)} + \frac {(a - b)}{ab^2 + bc^2 + ca^2}$
It is very éay to find $ N\geq\ 0$ so the proof is finished[/quote]
The inequality $ N \ge 0$ is not true, try $ c = 0,a = 1,b = 3$ ;)[/quote]
If we suppose c=max{a;b;c} ; I think Allnames's proof still right[/quote]
Sorry,in the truth, my proof has some mistakes,but with your one the proof is still wrong
Try it with$ a = o,b = 1,c = 1,1$

I have seen a similar problem:D
(Lê Trung Kien)
Given $ a, b, c \geq\ 0$. Prove that:
$ \frac{a}{b\plus{}c} \plus{}\frac{b}{c\plus{}a} \plus{}\frac{c}{a\plus{}b} \plus{} \frac{3(a^2b\plus{}b^2c\plus{}c^2a)}{ab^2\plus{}bc^2\plus{}ca^2} \geq\ 4$

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}
\]
[/quote]
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{a}{b\plus{}c}\minus{}\frac{1}{2}\right)\geq\frac{\sum(a^2c\minus{}a^2b)}{\sum a^2c}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{a\minus{}b\minus{}(c\minus{}a)}{2(b\plus{}c)}\geq\frac{(a\minus{}b)(b\minus{}c)(c\minus{}a)}{a^2c\plus{}b^2a\plus{}c^2b}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{a\minus{}b}{2}\left(\frac{1}{b\plus{}c}\minus{}\frac{1}{a\plus{}c}\right)\geq\frac{(a\minus{}b)(b\minus{}c)(c\minus{}a)}{a^2c\plus{}b^2a\plus{}c^2b}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)}\geq\frac{2(a\minus{}b)(b\minus{}c)(c\minus{}a)}{a^2c\plus{}b^2a\plus{}c^2b}.$
If $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\leq0$ then the inequality holds.
Let $ (a\minus{}b)(b\minus{}c)(c\minus{}a)>0$ and $ \frac{a^2c\plus{}b^2a\plus{}c^2b}{a^2b\plus{}b^2c\plus{}c^2a}\equal{}t.$ Then $ t>1.$
By AM-GM we obtain: $ \sum_{cyc}\frac{(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)}\geq3\sqrt[3]{\frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{(a\plus{}b)^2(a\plus{}c)^2(b\plus{}c)^2}}.$
Thus, it remains to prove that
 $ 27(a^2c\plus{}b^2a\plus{}c^2b)^3\geq8(a\plus{}b)^2(a\plus{}c)^2(b\plus{}c)^2(a\minus{}b)(b\minus{}c)(c\minus{}a).$
But $ (a\plus{}b)(a\plus{}c)(b\plus{}c)\equal{}\sum_{cyc}(a^2b\plus{}a^2c)\plus{}2abc\leq\frac{4}{3}\sum_{cyc}(a^2b\plus{}a^2c).$
Id est, it remains to prove that $ 27t^3\geq8\cdot\frac{16}{9}(t\plus{}1)^2(t\minus{}1),$ which obvious.

[quote="nguoivn"]I have seen a similar problem:D
(Lê Trung Kien)
Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3(a^2b \plus{} b^2c \plus{} c^2a)}{ab^2 \plus{} bc^2 \plus{} ca^2} \geq\ 4$[/quote]
not similar they are really different

Oh . I have found this way but I haven't setting : $ t \equal{} \frac {\sum a^2c}{\sum ac^2}$ and try to use GMV,but i can't :D
@Mathpro: But it's similar form :P

[quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}
\]
[/quote]
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {a}{b \plus{} c} \minus{} \frac {1}{2}\right)\geq\frac {\sum(a^2c \minus{} a^2b)}{\sum a^2c}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {a \minus{} b \minus{} (c \minus{} a)}{2(b \plus{} c)}\geq\frac {(a \minus{} b)(b \minus{} c)(c \minus{} a)}{a^2c \plus{} b^2a \plus{} c^2b}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {a \minus{} b}{2}\left(\frac {1}{b \plus{} c} \minus{} \frac {1}{a \plus{} c}\right)\geq\frac {(a \minus{} b)(b \minus{} c)(c \minus{} a)}{a^2c \plus{} b^2a \plus{} c^2b}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a \minus{} b)^2}{(a \plus{} c)(b \plus{} c)}\geq\frac {2(a \minus{} b)(b \minus{} c)(c \minus{} a)}{a^2c \plus{} b^2a \plus{} c^2b}.$
If $ (a \minus{} b)(b \minus{} c)(c \minus{} a)\leq0$ then the inequality holds.
Let $ (a \minus{} b)(b \minus{} c)(c \minus{} a) > 0$ and $ \frac {a^2c \plus{} b^2a \plus{} c^2b}{a^2b \plus{} b^2c \plus{} c^2a} \equal{} t.$ Then $ t > 1.$
By AM-GM we obtain: $ \sum_{cyc}\frac {(a \minus{} b)^2}{(a \plus{} c)(b \plus{} c)}\geq3\sqrt [3]{\frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2}}.$
Thus, it remains to prove that
 $ 27(a^2c \plus{} b^2a \plus{} c^2b)^3\geq8(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2(a \minus{} b)(b \minus{} c)(c \minus{} a).$
But $ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} \sum_{cyc}(a^2b \plus{} a^2c) \plus{} 2abc\leq\frac {4}{3}\sum_{cyc}(a^2b \plus{} a^2c).$
Id est, it remains to prove that $ 27t^3\geq8\cdot\frac {16}{9}(t \plus{} 1)^2(t \minus{} 1),$ which obvious.[/quote]
Thanks for you interest, arqady. :)
But I think your method doesn't work for the following
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} k\cdot \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge k \plus{} \frac {3}{2}
\]
for all $ a,b,c \ge 0$ and $ 0 \le k \le \frac {6}{5}$.
;)
[quote="nguoivn"]I have seen a similar problem:D
(Lê Trung Kien)
Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3(a^2b \plus{} b^2c \plus{} c^2a)}{ab^2 \plus{} bc^2 \plus{} ca^2} \geq\ 4$[/quote]
It is weaker than
\[ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \le \frac {4}{27}(a \plus{} b \plus{} c)^3
\]
;)

[quote="mitdac123"]When $ k \equal{} \frac {11}{5}$;$ c \equal{} 0$
<=>$ \frac {16}{5} \frac {a}{b} \plus{} \frac {b}{a} \ge \frac {11}{5} \plus{} \frac {3}{2}$ (Which is wrong :|)[/quote]
Sorry, mitdac123.
The correct is $ 0 \le k \le \frac {6}{5}$.
I have got a big mistake that $ 1.2 \equal{} \frac {11}{5}$  :blush:

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}
\]
[/quote]
asume $ p \equal{} 1$ and [b]Lemma[/b] $ ab^2 \plus{} bc^2 \plus{} ca^2\le \frac {4}{27} \minus{} r$

$ \Leftrightarrow \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {\sum ab(a\plus{}b)}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {7}{2}$

we have $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {\sum ab(a\plus{}b)}{ab^2 \plus{} bc^2 \plus{} ca^2}\ge \frac{1\minus{}2q\plus{}3r}{q\minus{}r}\plus{}\frac{27q\minus{}81r}{4\minus{}27r}$

we need prove that
$ \frac{1\minus{}2q\plus{}3r}{q\minus{}r}\plus{}\frac{27q\minus{}81r}{4\minus{}27r}\ge \frac {7}{2}$

$ \Leftrightarrow \frac{1\plus{}r}{q\minus{}r}\minus{}2\plus{}\frac{27q\minus{}12}{4\minus{}27r}\plus{}3\ge \frac {7}{2}$

$ \Leftrightarrow \frac {1 \plus{} r}{q \minus{} r} \plus{} \frac {27q \minus{} 12}{4 \minus{} 27r}\ge \frac {5}{2}$

$ \Leftrightarrow \minus{} 135r^2 \plus{} r(81q \plus{} 2) \plus{} (54q^2 \plus{} 8 \minus{} 44q)\ge 0$

$ f(r) \equal{} \minus{} 135r^2 \plus{} r(81q \plus{} 2) \plus{} (54q^2 \plus{} 8 \minus{} 44q)$

$ f'(r) \equal{} \minus{} 270r \plus{} 81q \plus{} 2 \ge 0$ (because $ q\ge 9r$)

$ \Rightarrow f(r)\ge f(\frac {4q \minus{} 1}{9}) \equal{} \frac {570q^2 \minus{} 349q \plus{} 55}{9}\ge 0$

A similar form is the following
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{}\frac{3abc}{2(a^2b\plus{}b^2c\plus{}c^2a)} \ge 2\]
for all $ a,b,c \ge 0$.
:)

[quote="can_hang2007"][quote="bunhiacovski"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}
\]
[/quote]
asume $ p \equal{} 1$ and [b]Lemma[/b] $ ab^2 \plus{} bc^2 \plus{} ca^2\le \frac {4}{27} \minus{} r$

$ \Leftrightarrow \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {\sum ab(a \plus{} b)}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {7}{2}$

we have $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {\sum ab(a \plus{} b)}{ab^2 \plus{} bc^2 \plus{} ca^2}\ge \frac {1 \minus{} 2q \plus{} 3r}{q \minus{} r} \plus{} \frac {27q \minus{} 81r}{4 \minus{} 27r}$

we need prove that
$ \frac {1 \minus{} 2q \plus{} 3r}{q \minus{} r} \plus{} \frac {27q \minus{} 81r}{4 \minus{} 27r}\ge \frac {7}{2}$

$ \Leftrightarrow \frac {1 \plus{} r}{q \minus{} r} \minus{} 2 \plus{} \frac {27q \minus{} 12}{4 \minus{} 27r} \plus{} 3\ge \frac {7}{2}$

$ \Leftrightarrow \frac {1 \plus{} r}{q \minus{} r} \plus{} \frac {27q \minus{} 12}{4 \minus{} 27r}\ge \frac {5}{2}$

$ \Leftrightarrow \minus{} 135r^2 \plus{} r(81q \plus{} 2) \plus{} (54q^2 \plus{} 8 \minus{} 44q)\ge 0$

$ f(r) \equal{} \minus{} 135r^2 \plus{} r(81q \plus{} 2) \plus{} (54q^2 \plus{} 8 \minus{} 44q)$

$ f'(r) \equal{} \minus{} 270r \plus{} 81q \plus{} 2 \ge 0$ (because $ q\ge 9r$)

$ \Rightarrow f(r)\ge f(\frac {4q \minus{} 1}{9}) \equal{} \frac {570q^2 \minus{} 349q \plus{} 55}{9}\ge 0$[/quote]
So when do we have equality? I don't see it in your proof?! ;)[/quote]

Here is this, Can.

When we transform it into $ p,q,r$ we get the inequality becomes

\[ \frac{{1 \minus{} 2q \plus{} 3r}}{{q \minus{} r}} \plus{} \frac{{27\left( {q \minus{} 3r} \right)}}{{4 \minus{} 27r}} \ge \frac{7}{2}
\]

If we setting $ f\left( r \right) \equal{} \frac{{1 \minus{} 2q \plus{} 3r}}{{q \minus{} r}} \plus{} \frac{{27\left( {q \minus{} 3r} \right)}}{{4 \minus{} 27r}}$ then
\[ f'\left( r \right) \equal{} \frac{{q \plus{} 1}}{{\left( {q \minus{} r} \right)^2 }} \plus{} \frac{{27q \minus{} 12}}{{\left( {4 \minus{} 27r} \right)^2 }} \ge \frac{{q \plus{} 1}}{{q^2 }} \plus{} \frac{{27q \minus{} 12}}{{16}} \ge 0
\]
By using Schur inequality for third degree we need to prove in two cases

$ r \ge 0$ with $ q \le \frac{1}{4}$

$ r \ge \frac{{4q \minus{} 1}}{9}$ with $ q \ge \frac{1}{4}$

In the first case we need to show that
\[ \frac{{1 \minus{} 2q}}{q} \plus{} \frac{{27q}}{4} \ge \frac{7}{2}
\]
\[ \Leftrightarrow 54q^2  \minus{} 44q \plus{} 8 \ge 0
\]
(true).

In the second case we need to show that
\[ \frac{{1 \minus{} 2q \plus{} \frac{{4q \minus{} 1}}{3}}}{{q \minus{} \frac{{4q \minus{} 1}}{9}}} \plus{} \frac{{27\left( {q \minus{} \frac{{4q \minus{} 1}}{3}} \right)}}{{4 \minus{} 3\left( {4q \minus{} 1} \right)}} \ge \frac{7}{2}
\]
\[ \Leftrightarrow \frac{{6\left( {1 \minus{} q} \right)}}{{5q \plus{} 1}} \plus{} \frac{{9\left( {1 \minus{} q} \right)}}{{7 \minus{} 12q}} \ge \frac{7}{2}
\]
\[ \Leftrightarrow \frac{{\left( {3q \minus{} 1} \right)\left( {158q \minus{} 53} \right)}}{{2\left( {5q \plus{} 1} \right)\left( {7 \minus{} 12q} \right)}} \ge 0
\]
which is obviously true.
The inequality has been proved. Equality holds if and only if $ a\equal{}b\equal{}c$.

[quote="mitdac123"]*if
\[ a,b,c
\]
be side lengths of a triangle.
Assume :$ p \equal{} 1$
Use:
+$ \sum 3(a^2b ) \ge 2\sum ab^2 \plus{} 3abc$
+$ r \ge \frac {5q \minus{} 1}{18}$

$ LHS \minus{} RHS \ge \sum \frac {a}{b \plus{} c} \plus{} \frac {2abc}{\sum ab(a \plus{} b) \minus{} 2abc} \minus{} 2 \equal{} \frac {q \minus{} 5r \minus{} 4q^2 \plus{} 27rq \minus{} 27r^2}{(q \minus{} r)(q \minus{} 5r)} \ge \frac {(1 \minus{} 3q)(7 \minus{} 17q)}{36(q \minus{} r)(q \minus{} 5r)} \ge 0$
*if
\[ a,b,c
\]
won't be  side lengths of a triangle (Are there some mistakes in gammar :( ? )
[hide]I will think about it tomorrow .Now , I must sleep :P [/hide]
$ LHS \ge \sum \frac {a}{b \plus{} c} \ge \sum 1 \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)} > 2$
Gernalize the first ineq .
$ a,b,c \ge 0;k \ge 0$
$ \sum \frac {a}{b \plus{} c} \plus{} k\frac {\sum a^2b}{\sum ab^2} \ge min( \frac {3}{2} \plus{} k,2\sqrt {k \plus{} 1} )$
(it's easy to prove it  by GMV)[/quote]

Notice the second equality case $ a\equal{}b,c\equal{}0$ for a good solution.

Sorry ,my solution was worng:p
@lọc: I know that case :|

[quote="can_hang2007"]
But I think your method doesn't work for the following
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} k\cdot \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge k \plus{} \frac {3}{2}
\]
for all $ a,b,c \ge 0$ and $ 0 \le k \le \frac {6}{5}$.
;)
[/quote]
But it works for all positive $ a,$ $ b$ and $ c$ and $ 0 \le k \le \frac {9\sqrt[3]9}{16}.$

[quote="nguoivn"]I have seen a similar problem:D
(Lê Trung Kien)
Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3(a^2b \plus{} b^2c \plus{} c^2a)}{ab^2 \plus{} bc^2 \plus{} ca^2} \geq\ 4$[/quote]
Here is 2 solutions of mine...

Dear can_hang2007,

the last line of your nice proof is

$ 13\minus{}93q\plus{}270q^2\minus{}108r$

but I think it should be

$ 13\plus{}15q\plus{}270q^2\minus{}216r$   :wink:

OK, dear can_hang2007. I will check again  :wink:

[quote="Honey_S"][quote="bunhiacovski"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}
\]
[/quote]
asume $ p \equal{} 1$ and [b]Lemma[/b] $ ab^2 \plus{} bc^2 \plus{} ca^2\le \frac {4}{27} \minus{} r$

$ \Leftrightarrow \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {\sum ab(a \plus{} b)}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {7}{2}$

we have $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {\sum ab(a \plus{} b)}{ab^2 \plus{} bc^2 \plus{} ca^2}\ge \frac {1 \minus{} 2q \plus{} 3r}{q \minus{} r} \plus{} \frac {27q \minus{} 81r}{4 \minus{} 27r}$

we need prove that
$ \frac {1 \minus{} 2q \plus{} 3r}{q \minus{} r} \plus{} \frac {27q \minus{} 81r}{4 \minus{} 27r}\ge \frac {7}{2}$

$ \Leftrightarrow \frac {1 \plus{} r}{q \minus{} r} \minus{} 2 \plus{} \frac {27q \minus{} 12}{4 \minus{} 27r} \plus{} 3\ge \frac {7}{2}$

$ \Leftrightarrow \frac {1 \plus{} r}{q \minus{} r} \plus{} \frac {27q \minus{} 12}{4 \minus{} 27r}\ge \frac {5}{2}$

$ \Leftrightarrow \minus{} 135r^2 \plus{} r(81q \plus{} 2) \plus{} (54q^2 \plus{} 8 \minus{} 44q)\ge 0$

$ f(r) \equal{} \minus{} 135r^2 \plus{} r(81q \plus{} 2) \plus{} (54q^2 \plus{} 8 \minus{} 44q)$

$ f'(r) \equal{} \minus{} 270r \plus{} 81q \plus{} 2 \ge 0$ (because $ q\ge 9r$)

$ \Rightarrow f(r)\ge f(\frac {4q \minus{} 1}{9}) \equal{} \frac {570q^2 \minus{} 349q \plus{} 55}{9}\ge 0$[/quote]

It is also my proof.  :wink:[/quote]

I think $ f(r)$ is wrong.

I checked many times your computations and I found always

$ f(r) \equal{} \minus{} 189r^2 \plus{} r(81q \minus{} 2) \plus{} (54q^2 \plus{} 8 \minus{} 44q)$

so, I think your proof doesn't work with this $ f(r)$.

Can you please tell me if I am right?  :wink:

Thank you very much

yes, this my proof is wrong because it isn't "=" occur  :)
I had another proof for it

[quote="can_hang2007"][quote="manlio"]Dear can_hang2007,

the last line of your nice proof is

$ 13 \minus{} 93q \plus{} 270q^2 \minus{} 108r$

but I think it should be

$ 13 \plus{} 15q \plus{} 270q^2 \minus{} 216r$   :wink:[/quote]
I dont think so, manlio. Please check your calculation ;)[/quote]

You are right, can_hang2007. Sorry  :blush:  :blush:

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that

\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}.\][/quote]$ \sum_{sym}{\frac {a}{b \plus{} c}} \plus{} \frac { b^2c \plus{} c^2a \plus{} a^2b}{ bc^2 \plus{} ca^2 \plus{} ab^2 } \minus{} \frac {5}{2}$

$ \equal{} \sum_{cyc}{\frac {a(b \minus{} c)^2[2(5a \plus{} 8b)c^2 \minus{} (7a^2 \minus{} ab \plus{} 3b^2)c \plus{} 2a^3 \plus{} 4a^2b \minus{} 11ab^2 \plus{} 6b^3]}{6(b \plus{} c)(c \plus{} a)(a \plus{} b)(ab^2 \plus{} bc^2 \plus{} ca^2)}}$

$ \geq0,$

which is true for

$ 8(5a \plus{} 8b)(2a^3 \plus{} 4a^2b \minus{} 11ab^2 \plus{} 6b^3) \minus{} (7a^2 \minus{} ab \plus{} 3b^2)^2$

$ \equal{} 31a^4 \plus{} 302a^3b \minus{} 227a^2b^2 \minus{} 458ab^3 \plus{} 375b^4$

$ \equal{} \frac {(31a \minus{} 29b)^2(14a^2 \plus{} 161ab \plus{} 186b^2) \plus{} b(1519a^3 \plus{} 440a^2b \plus{} 255ab^2 \plus{} 6324b^3)}{434}$

$ \geq0.$

[b]Generalization[/b] If $ a,b,c$ be nonnegative real numbers such that $ bc \plus{} ca \plus{} ab > 0,$ then

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} k\cdot \frac { b^2c \plus{} c^2a \plus{} a^2b}{ bc^2 \plus{} ca^2 \plus{} ab^2 }\geq\frac {3}{2} \plus{} k$

holds if and only if $ \frac {1}{2} \minus{} \sqrt {2}\leq k\leq k_1 \equal{} 1.90128\cdots$,

where $ k_1$ is the greatest real root of the following irreducible polynomial over $ \mathbb{Q}:$

$ 268435456k^{10} \plus{} 2281701376k^9 \plus{} 8034451456k^8 \plus{} 12571639808k^7 \minus{} 932267008k^6$

$ \minus{} 43105478656k^5 \minus{} 92405285760k^4 \minus{} 107561411280k^3 \minus{} 76942450476k^2$

$ \minus{} 32259764862k \minus{} 6253684947;$

with equality if $ k \equal{} k_1,a \equal{} x_5 \equal{} 0.0256718\cdots,b \equal{} x_3 \equal{} 0.360546\cdots,c \equal{} x_2 \equal{} 0.613781\cdots,$

where $ x_2,x_3$ and $ x_5$ are three real roots (the second, the third and the fifth respectively) of the following irreducible polynomial over $ \mathbb{Q}:$

$ 1259712x^{30} \minus{} 12597120x^{29} \plus{} 33382368x^{28} \plus{} 40205808x^{27} \minus{} 320596704x^{26}$

$ \plus{} 568698732x^{25} \minus{} 995429817x^{24} \plus{} 3282771564x^{23} \minus{} 8511820767x^{22}$

$ \plus{} 16329623058x^{21} \minus{} 29728112184x^{20} \plus{} 53377049466x^{19} \minus{} 82591209360x^{18}$

$ \plus{} 105163867098x^{17} \minus{} 116564168019x^{16} \plus{} 120616149122x^{15} \minus{} 116723994616x^{14}$

$ \plus{} 99747213078x^{13} \minus{} 71302856054x^{12} \plus{} 41502868742x^{11} \minus{} 19573294802x^{10}$

$ \plus{} 7580967178x^{9} \minus{} 2499870437x^{8} \plus{} 741218666x^{7} \minus{} 204119097x^{6} \plus{} 50142336x^{5}$

$ \minus{} 9972698x^{4} \plus{} 1441976x^{3} \minus{} 134510x^{2} \plus{} 6664x \minus{} 103.$

[b]Remark[/b] $ k \equal{} \frac {1}{2} \minus{} \sqrt {2}$ see here : http://www.mathlinks.ro/viewtopic.php?t=257809

[quote="Ji Chen"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a}{ab^2 \plus{} bc^2 \plus{} ca^2} \ge \frac {5}{2}.
\]
[/quote]$ \sum_{sym}{\frac {a}{b \plus{} c}} \plus{} \frac { b^2c \plus{} c^2a \plus{} a^2b}{ bc^2 \plus{} ca^2 \plus{} ab^2 } \minus{} \frac {5}{2}$

$ \equal{} \sum_{cyc}{\frac {a(b \minus{} c)^2[2(5a \plus{} 8b)c^2 \minus{} (7a^2 \minus{} ab \plus{} 3b^2)c \plus{} 2a^3 \plus{} 4a^2b \minus{} 11ab^2 \plus{} 6b^3]}{6(b \plus{} c)(c \plus{} a)(a \plus{} b)(ab^2 \plus{} bc^2 \plus{} ca^2)}}$

$ \geq0,$

which is true for

$ 8(5a \plus{} 8b)(2a^3 \plus{} 4a^2b \minus{} 11ab^2 \plus{} 6b^3) \minus{} (7a^2 \minus{} ab \plus{} 3b^2)^2$

$ \equal{} 31a^4 \plus{} 302a^3b \minus{} 227a^2b^2 \minus{} 458ab^3 \plus{} 375b^4$

$ \equal{} \frac {(31a \minus{} 29b)^2(14a^2 \plus{} 161ab \plus{} 186b^2) \plus{} b(1519a^3 \plus{} 440a^2b \plus{} 255ab^2 \plus{} 6324b^3)}{434}$

$ \geq0.$

[b]Generalization[/b] If $ a,b,c$ be nonnegative real numbers such that $ bc \plus{} ca \plus{} ab > 0,$ then

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} k\cdot \frac { b^2c \plus{} c^2a \plus{} a^2b}{ bc^2 \plus{} ca^2 \plus{} ab^2 }\geq\frac {3}{2} \plus{} k$

holds if and only if $ \frac {1}{2} \minus{} \sqrt {2}\leq k\leq k_1 \equal{} 1.90128\cdots$,

where $ k_1$ is the greatest real root of the following irreducible polynomial over $ \mathbb{Q}:$

$ 268435456k^{10} \plus{} 2281701376k^9 \plus{} 8034451456k^8 \plus{} 12571639808k^7 \minus{} 932267008k^6$

$ \minus{} 43105478656k^5 \minus{} 92405285760k^4 \minus{} 107561411280k^3 \minus{} 76942450476k^2$

$ \minus{} 32259764862k \minus{} 6253684947;$

with equality if $ k \equal{} k_1,a \equal{} x_5 \equal{} 0.0256718\cdots,b \equal{} x_3 \equal{} 0.360546\cdots,c \equal{} x_2 \equal{} 0.613781\cdots,$

where $ x_2,x_3$ and $ x_5$ are three real roots (the second, the third and the fifth respectively) of the following irreducible polynomial over $ \mathbb{Q}:$

$ 1259712x^{30} \minus{} 12597120x^{29} \plus{} 33382368x^{28} \plus{} 40205808x^{27} \minus{} 320596704x^{26}$

$ \plus{} 568698732x^{25} \minus{} 995429817x^{24} \plus{} 3282771564x^{23} \minus{} 8511820767x^{22}$

$ \plus{} 16329623058x^{21} \minus{} 29728112184x^{20} \plus{} 53377049466x^{19} \minus{} 82591209360x^{18}$

$ \plus{} 105163867098x^{17} \minus{} 116564168019x^{16} \plus{} 120616149122x^{15} \minus{} 116723994616x^{14}$

$ \plus{} 99747213078x^{13} \minus{} 71302856054x^{12} \plus{} 41502868742x^{11} \minus{} 19573294802x^{10}$

$ \plus{} 7580967178x^{9} \minus{} 2499870437x^{8} \plus{} 741218666x^{7} \minus{} 204119097x^{6} \plus{} 50142336x^{5}$

$ \minus{} 9972698x^{4} \plus{} 1441976x^{3} \minus{} 134510x^{2} \plus{} 6664x \minus{} 103.$

[b]Remark[/b] $ k \equal{} \frac {1}{2} \minus{} \sqrt {2}$ see here : http://www.mathlinks.ro/viewtopic.php?t=257809[/quote]

Is it bottema or chen86 ?


hi,

as the title suggests i am trying to find "a".

i have the following equation;

(a/v) - (( a - ( a - ((( a - c ) * y ) + z )))) = n

i have got to the following (dont know if it is correct or not) but i am still stuck trying to find "a";

v(z-n+yc) = vya + a

any help would be appreciated!!!

thanks

hello, do you mean the equation
$ \frac{a}{v}\minus{}\left(a\minus{}(a\minus{}c)y\plus{}z\right)\equal{}n$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean the equation
$ \frac {a}{v} \minus{} \left(a \minus{} (a \minus{} c)y \plus{} z\right) \equal{} n$?
Sonnhard.[/quote]

i'm sorry, i've not done this for 10 years.

i think that it should be;

$ \frac {a}{v} \minus{} \left(a \minus{} (a \minus{} ((a \minus{} c)y \plus{} z\right))) \equal{} n$

thanks

hello, simplifying your equation step by step you will get
$ \frac{a}{v}\minus{}(a\minus{}(a\minus{}(a\minus{}c)y\minus{}z))\equal{}n$
$ \frac{a}{v}\minus{}(a\minus{}a\plus{}(a\minus{}c)y\plus{}z)\equal{}n$
$ \frac{a}{v}\minus{}(a\minus{}c)y\minus{}z\equal{}n$
$ \frac{a}{v}\minus{}ay\plus{}cy\minus{}z\equal{}n$
$ a\left(\frac{1}{v}\minus{}y\right)\equal{}n\minus{}cy\plus{}z$
$ a\frac{(1\minus{}vy)}{v}\equal{}n\minus{}cy\plus{}z$
$ a\equal{}\frac{(n\minus{}cy\plus{}z)v}{1\minus{}vy}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, simplifying your equation step by step you will get
$ \frac {a}{v} \minus{} (a \minus{} (a \minus{} (a \minus{} c)y \minus{} z)) \equal{} n$
$ \frac {a}{v} \minus{} (a \minus{} a \plus{} (a \minus{} c)y \plus{} z) \equal{} n$
$ \frac {a}{v} \minus{} (a \minus{} c)y \minus{} z \equal{} n$
$ \frac {a}{v} \minus{} ay \plus{} cy \minus{} z \equal{} n$
$ a\left(\frac {1}{v} \minus{} y\right) \equal{} n \minus{} cy \plus{} z$
$ a\frac {(1 \minus{} vy)}{v} \equal{} n \minus{} cy \plus{} z$
$ a \equal{} \frac {(n \minus{} cy \plus{} z)v}{1 \minus{} vy}$.
Sonnhard.[/quote]

i am sorry, i presented this incorrectly.

the original equation should have been;

$ \frac {a}{v} \minus{} a \minus{} a \minus{} (a \minus{} c)y \minus{} z \equal{} n$

hello, in this case you will get
$ \frac{a}{v}\minus{}2a\minus{}ay\equal{}n\plus{}z\minus{}cy$
$ a\left(\frac{1}{v}\minus{}2\minus{}y\right)\equal{}n\plus{}z\minus{}cy$
$ a\frac{(1\minus{}2v\minus{}yv)}{v}\equal{}n\plus{}z\minus{}cy$
$ a\equal{}\frac{v(n\plus{}z\minus{}cy)}{1\minus{}2v\minus{}yv}$.
The denominator must be not equal zero.
Sonnhard.


Nine circular spheres are packed inside a unit cube in such a way that one of them has its center at the center of the cube and each other of the others is tangent to the center sphere and to three faces of the cube. what is the radius of each sphere?

I have no idea :huh:

[hide]

Okay, so we have one sphere in the middle, and 4 on the top/ 4 on the bottom. Therefore, all we have to do is find the diagonal of the unitcube and divide by $ 2\cdot3=6$ to get the radius:

$ \frac{\sqrt{\sqrt{1^{2}+1^{2}}+1^{2}}}{6}$
$ =\frac{\sqrt{\sqrt{2}+1}}{6}$
$ =\frac{\sqrt{2.414213562}}{6}$
$ =\frac{1.553773974}{6}$
$ =.258962329$

Hence our answer is: $ \boxed{.258962329}$

[/hide]

-I hope this helps!

If you've already solved that problem, try this, too.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155250[/url]

Nexmus, your solution is incorrect.

Yes you do have to find the diagonal of the cube, but you do not divide by 6.

Visualize it carefully. When you cut the cube by a diagonal, you get a $ 1 \text{ by }\sqrt{2}$ box with 5 cubes inside. However (this is the tricky part) the balls on the sides of the box do NOT touch the box. The balls are only tangent to the top and bottom of the box, not the sides.

To make this easier to understand, look at the box from the top. at the corners, there is a bit of empty space. When the diagonal is cut, this is the space that seperates the box from the balls.

So the key to this problem was realizing this and cutting the box at a diagonal. Once that is done, assign some variables and a little bit of Pythagoreas should bring you to your answer.

Examine the center cube and one corner cube. The distance from the center of the corner cube to the closest cube-corner is sqrt(1+1+1)*r = sqrt(3)*r, where r is the radius of the sphere.
Drawing the diagonal of the cube, which is sqrt(1+1+1) = sqrt(3), we see that the diagonal contains 4 sphere radii and two lengths of sqrt(3)*r.

So sqrt(3) = (4+2sqrt(3))r and r = 2sqrt(3)-3.

Riders, I dont think thats the right answer.

I agree. Your final simplification was faulty and the final answer is

$ \frac{2\sqrt3-3}{2}$

oh yeah...sorry I forgot the /2 at the end.


conics
parabola
algebra
function
domain
inequalities proposed
inequalities

Move everything to the L.H.S. it's convex w.r.t $ a$ since it's an upwards facing parabola or a line, so the max occurs at one of the endpoints of the domain of a, which is 0 or 1. Similarly for b and c.

So we only need to check a,b,c when they are all either 0 or 1. This is trivial.

[quote="rofler"]Move everything to the L.H.S. it's convex w.r.t since it's an upwards facing parabola or a line, so the max occurs at one of the endpoints of the domain of a, which is 0 or 1. Similarly for b and c.

So we only need to check a,b,c when they are all either 0 or 1. This is trivial.[/quote]
Sorry, Can U post full proof ? And I warn to say grade 8 only study AM-GM & C-S

[quote="onlylove_math"]Let $ 0 \le\ a;b;c \le\ 1$. Prove that :

$ a^2 \plus{} b^2 \plus{} c^2 \le\ 1 \plus{} a^2b \plus{} b^2c \plus{} c^2a$[/quote]
Ok,guy
$ (1 \minus{} a)(1 \minus{} b)a\ge 0$thus $ a^2b \plus{} a\ge a^2 \plus{} ab$
It is similar to $ b,c$
And $ (1 \minus{} a)1 \minus{} b)(1 \minus{} c)\ge 0$ hence $ 1 \plus{} \sum ab\ge \sum a \plus{} abc$ we have a stronger ineq
$ a^2 \plus{} b^2 \plus{} c^2\plus{}abc \le\ 1 \plus{} a^2b \plus{} b^2c \plus{} c^2a \plus{}$.Ok,my friend
With the proof which be used convex,we only need a 2-degree  polyminal is a convex,and it is max (pr min) at $ 0$ or $ 1$ At all,it is not suitable to grade8
----------------------------------------------------------------------------------------------------------------------
P/S:Oh,your avatar!!!!!!!!! :o  a beautiful girl,tell me more about her,I quite like her,very beautiful girl
Any her photo ?

I say that some grade 8's must know about convex. Otherwise, you just show that either:

$ (1\minus{}b)a^2\minus{}c^2a\plus{}b^2\plus{}c^2\minus{}b^2c \le b^2\plus{}c^2\minus{}b^2c$ <--> $ (1\minus{}b)a^2\minus{}c^2a \le 0$

or $ (1\minus{}b)a^2\minus{}c^2a\plus{}b^2\plus{}c^2\minus{}b^2c \le 1\minus{}b\minus{}c^2\plus{}b^2\plus{}c^2\minus{}b^2c$ <--> $ (1\minus{}b)a^2\minus{}c^2a \le 1\minus{}b\minus{}c^2$

Now, if $ (1\minus{}b)a > c^2$ and $ (1\minus{}b)(a^2\minus{}1)\minus{}c^2(a\minus{}1) >0$ then a,b are not equal to 1, so we have the second one can be rewritten as $ (1\minus{}b)(a\plus{}1)\minus{}c^2 < 0$ so $ (1\minus{}b)(a\plus{}1) < c^2$ contradicting the first one.

Therefore, either $ (a,b,c) \le (0,b,c)$ or $ (a,b,c) \le (1,b,c)$ so then applying the same logic to b,c we're done.

Besides, if you're doing math contests, you know what you know, regardless of whether they explicitly teach you it or not.


MATHCOUNTS
trigonometry
AMC
AIME
logarithms
AMC 10
Putnam

what subjects stay the same...like number theory, i guess
and what subjects is there a more extent of...like geo or alg??
How much harder is it than MC??
What new concepts??
idk what else, just recommendations and opinions i guess :P

Um, each subject is extended, and on AMC 12/AIME/everything else, there is trig and logs

Basically, MATHCOUNTS is a middle school test and the AMCs and AIMES are high school tests. Even though MC could get really hard, it doesn't ask about anything beyond middle school level. For example, I see far less (complicated) algebra problems at MC than on the AMCs. Also, like AIME15 said, the AMCs and AIMEs expect you to know how to use trig and logarithms and complex numbers and that sort.

In my opinion: AMCs are easier than MC. Nats is pretty beast :D. Although, Mathcounts is rather more fun to do.

If you didn't already know, you should try some old tests to find out for yourself.
http://www.artofproblemsolving.com/Wiki/index.php/AMC_10_Problems_and_Solutions

I disagree. In my opinion, Mc Nats is around an AMC10 level. The AMC12/AIME are much harder.

[quote="Ihatepie"]Mc Nats is around an AMC10 level.[/quote]

I see this comparison a lot, and I really have to disagree with it. MATHCOUNTS has a totally different type of problems, and while the concepts may be around the same level, their applications in MATHCOUNTS and AMC10 are completely different. And the biggest part is probably the speed aspect...MATHCOUNTS is more or less the most speed intensive (by a lot) competition out there that people care about.

I'd have to agree with the speed part.

Now, I've always thought that completing a national mathcounts competition [b]and perfecting[/b] it was far more difficult than perfecting an AMC 10, which is not very hard for me to do. However, I havent thought very much about whether it was the problems themselves or the time. It may very well be that the lack of time makes the last few problems on a Nationals test seem much harder than on an AMC 10. However, I think that the very last few sprints, targets, and teams are harder than the AMC 10.

For the most part though, I'd have to agree with math154.

Comparing AMC and MathCounts is often comparison that should not be done. Why?

They are very different.

There should be no discrepancy on which one's harder. AMC is by far harder than MathCounts. Let's face it: AMC is ultimately a test to represent United States in international math contest. MathCounts is an exam for middle schoolers within United States. So, you see the difference in the size of game.

I could go on but I can guarantee that no one who has done both AMC and MathCounts will come out and say MathCounts is harder. That's almost as bad as saying Putnam is comparable to AMC 12.

I haven't heart that one yet but you never know in these days.

[quote="Silverfalcon"]I could go on but I can guarantee that no one who has done both AMC and MathCounts will come out and say MathCounts is harder. [/quote]

Uh hi? I scored a 21 in nats sprint (barely finished) and a 138.0 on the AMC 10 in 8th grade (which i finished in 20 minutes :| checking work just DOESN'T WORK FOR ME )

21/30=7/10<23/25 :(

nice comparision of points pi-dimension...anways, where can I learn about complex numbers, trig functions and logs briefly? like site or books or something :?:  :surrender:

Volume 2 is [b]very[/b] good if you want a good introduction to those topics. After you get the basics, do lots of problems.

Also, I am not sure if Silverfalcon is referring to the AMC 12 or AMC 10. I would immediately say that the AMC 10 is less difficult than National Mathcounts. The AMC 12 can [b]definitely[/b] not be compared to Mathcounts; it has topics that would not appear in Mathcounts.

Also my scores were  not as impressive as PI-Dimension's, but I had a 129 on the 10A (21 right, 2 blank, 2 incorrect) and a 19 on the National Sprint. Also, I actually had 17/20 of the first twenty sprints, but only 2/10 of the last ten. However on the AMC 10, both of the ones I got wrong were below question 20 (I believe they were like 6 and 16 or so). That should probably show that I should've had a higher score on the AMC 10 as I missed the relatively easy problems. On the other hand, the Nationals score means I was fairly good on the easy to medium difficulty, but was not very good at the harder problems.

Make of this what you want.

I meant to compare AMC 10/AIME and ALL levels of Mathcounts. That's what I presumed the thread to be based on the title.

Ah. My mistake.

Well if you're comparing all levels of Mathcounts, there's no doubt that the AMC 10 is more difficult than Chapter. I would also say that it is more difficult than most problems at state, except for possibly the very last few on sprint, target, and team. Then again, it depends on whether the state test is relative difficult or relatively easy.

The best way to see how the AMCs are is to take practice tests, as Textangle said, which can be found either on the Wiki or the Resource section. People have different opinions.

i think national mathcounts is easier than amc 10 however i cant really make the comparison since i never made it to nats. and in practice everything was eashowever, when i was in 8th grade i got a ~117 [cant quite remember exactly] and got a 15 on the state sprint [yeah, i screwed up bad. ] ><
but on the amc, you only have 25 problems to do [with multiple choice] and 75 minutes compared to 30 problems and 40 minutes for mathcounts. plus, the problems at the end can get rather difficult, and you dont have multiple choice answers...


function
calculus
limit
floor function
integration
probability
expected value

Take a looped rope of length $1$. If you cut the rope independently and randomly in $n$ places, one will get $n$ smaller pieces of rope.
Determine the avergae size of the biggest piece as function of $n$.

Well i suppose it is not asked for the average length, but more or less the expected value of the maximum length. this is $E= \sum_i p(s_i)\cdot s_i$ where $s_i$ is a possible maximum length and $p(s_i)$ its properbility.

since we have infinite many possible maximum lengths, we need to use some calculus (as far as i know ;)  please correct me ...)

let us first take an example with $n=1$ ($f(0)$ is clearly $1$), because we can see the solution of this without calculus as well ( $f(1)=\frac34$, its the mid between the lowest possible maximum length ($\frac12$) and the hightest ($1$) ):

$\displaystyle E = \lim_{x\to\infty} \sum_{i=0}^x \frac1x\cdot \max(\frac ix, 1-\frac ix)$
(here shall $x$ be the number of possible maximum lengths)
$\displaystyle= \lim_{x\to\infty} \left( \sum_{i=0}^{\left\lfloor\frac x2 \right\rfloor} \frac1x\left( 1 - \frac ix\right) + \sum_{i=\left\lfloor\frac x2 \right\rfloor +1}^{x} \frac i{x^2} \right)$
(here we use a trick or let us call it simply the definition of a definite integral, where $t: =\frac ix$ and thus $dt= \frac1x$)
$\displaystyle= \int_0^{\frac12} (1-t)\,dt + \int_{\frac12}^1 t\,dt$
$\displaystyle= \left[t-\frac12t^2\right]_0^{\frac12} + \left[\frac12t^2\right]_{\frac12}^1$
$\displaystyle=\frac12 -\frac18 + \frac12 -\frac 18 = \boxed{\frac34}$

so, i am just noticing, that for $n\ge 2$ the problem become (at least with alculus) way more complicated, so i will first solve it für 2 cuts and soon and post a general solution latter ;)


10 chairs are evenly spaced around a round table and numbered clockwise from 1 through 10. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or directly across from his or her spouse. How many seating arrangements are possible?

[hide]First, let the men be A, B, C, D, E and let their spouses be M, N, O, P, S respectively.
Since the table is round, rotating the table does not produce distinct seating arrangements, so suppose we fix male A at the top seat. The other 4 males must be seated in the seats colored red (see diagram), giving 24 arrangements for the men.
Since the number of arrangements for the women is independent of the actual seating arrangement, we can say WLOG, that the men are seated as they are in the diagram (counterclockwise). A's spouse, M, can not sit in the graan seats because of the restrictions, meaning it must sit in a blue seat. In the same way, each man's spouse can only sit in 2 seats- the 2 seats 3 seats away from her spouse's seat in either direction.
No matter which seat M sits in, all the other women's seats will be fixed, so there are only 2 arrangements for the women.
This means there are a total of $ 24 \cdot 2 \equal{} \boxed{48}$ total seating arrangements.
[/hide]

[hide="Hint"]
Since the chairs are evenly spaced, $ \frac{360^{\circ}}{10\text{ chairs}} \equal{} 36^{\circ}\text{/ chair}$.

Further, $ 180^{\circ}\div 36^{\circ}\text{/ chair} \equal{} 5\text{ chairs}$.

No one is allowed to sit exactly five chairs away from their own spouse.
[/hide]


I was reading a calculus book and I realized that most of my textbooks define fractional exponents as:

$x^{y/z}$...however when I asked my math teacher she says the correct form is $x^{\frac{y}{z}}$...

I like writing them as $x^{y/z}$ because it's easier on my paper as well as computations but what do you say?

[quote="max_tm"]I think anybody will understand if you write $x/y$ as opposed to $\frac{x}{y}$ but personally I like the later because for some reason if I have to start cancelling like terms and such I find the former very messy.[/quote]

Yeah I agree but the reason why i like $x/y$ better is because i feel it is faster to convert to a radical...for example

$x^{2/3}= \sqrt[3]{x^{2}}$...the $\frac{x}{y}$ way takes like half a second longer thou  :lol:


Find all positive integers $ a,b,c,d$ such that $ a \plus{} b \plus{} d^2 \equal{} 4abc$.

No solution.
$ d^2 \plus{} b \equal{} a(4bc \minus{} 1)$. Set $ b \equal{} 2^z.t$. Then $ 1 \equal{} \left( {\frac{{ \minus{} b}}{{4bc \minus{} 1}}} \right) \equal{} {\left( {\frac{2}{{4bc \minus{} 1}}} \right)^z}\left( {\frac{{ \minus{} 1}}{{4bc \minus{} 1}}} \right)\left( {\frac{t}{{4bc \minus{} 1}}} \right) \equal{} {( \minus{} 1)^{\frac{{{{(4bc \minus{} 1)}^2} \minus{} 1}}{8}.z}}.( \minus{} 1).{\left( {\frac{{ \minus{} 1}}{t}} \right)^2} \equal{}  \minus{} 1$.

[quote="tronghieu"]No solution.
$ d^2 \plus{} b \equal{} a(4bc \minus{} 1)$. Set $ b \equal{} 2^z.t$. Then $ 1 \equal{} \left( {\frac {{ \minus{} b}}{{4bc \minus{} 1}}} \right) \equal{} {\left( {\frac {2}{{4bc \minus{} 1}}} \right)^z}\left( {\frac {t}{{4bc \minus{} 1}}} \right) \equal{} \minus{} 1$.[/quote]

Dear tronghieu, could you make it clearer how do you use Jacobi Symbol in solving this equation?

One soln without Legendre's symbol in generalized form?
(I think that Legendre's symbol for primes (not for generalized form) is more necessary in Olympiad contest  :wink: )


induction
inequalities
inequalities solved
144

Prove that: sqrt(2n+1)-sqrt(2n)+sqrt(2n-1)-..-sqrt(2)+1>sqrt((n+1)/2) for any n natural.

looking at this one I realize that this one is pure induction!
cheers mate!

I know it is induction because that is also what the hint in book says! But I don't know how!

Okay.. this is a very simple example of induction!
Solution:

n=0 we get 1>0.5 (true)
now we consider the inequality:
p(n) : sqrt(2n+1)-sqrt(2n)+sqrt(2n-1)-..-sqrt(2)+1>sqrt((n+1)/2) and we will try to prove that p(n+1) is true!
we add sqrt(2n+3)-sqrt(2n+2) in both members in p(n)
I will show that sqrt(2n+3)-sqrt(2n+2)+sqrt((n+1)/2)>sqrt((n+2)/2) =>
sqrt(2n+3)>sqrt((n+3)/2)+sqrt((n+1)/2) and we get 2n+3>(2n+3)/2+sqrt((n+2)(n+1)) and we get that (2n+3)/2>sqrt((n+2)(n+1)) => 4n^2+12n+9>4(n^2+3n+2) which is true!

cheers!


There is an island with a certain number of inhabitants, 10 of whom have a red face. For some reasons, a red face problem is a topic that is never-to-talk on this island. No mirrors or anything on the island, and nobody is allowed to say to the person who has a red face that the person has a red face. And law of island says that if a person learns that he has a red face, he have to commit a suicide at 11:00pm that day. Since the topic is never-to-talk, there are no ways for red faced people to realize that he has a red face, so everyone lives happily until an alien visits the island. Right before leaving the island, an alien tells everyone: “There are some people on this island who have a red face! You should all start thinking about it now.” Ten days later (that is, on the the tenth 11:00pm after this declaration), all ten red-faced inhabitants commit suicide.

Question: How did they realize that they had a red face?
What did an alien say to the islanders that they didn’t know already?
What was an alien's contribution?

Note: It might not be recursion problem.

This should probably be in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=337]Puzzles and Brainteasers[/url] forum. I'm coming up with a solution...

Could you tell me the a concise solution..

Basically if you hear that there is at least one person with red face and you look around and see no one with red face you kill yourself right away.

If you see one person with redface on the first day, and by the second day they havent killed themself then they must see you with redface, so then you know you have redface and must kill yourself on the second day.

If you see 2 people with redface and they dont kill themselves by the 2nd day,then they must also see you with redface so you must kill yourself on the third day....

etc. etc.

This is a way cool problem (a whole chapter is devoted to it in th book Math Hysteria by Ian Stewart).  To figure it out, start out with small cases (like an island of 2 people, for example) and work your way up like a recursion.


[quote="boxedexe"]Hi,

What are the 3 solutions for solving 4a=a^3.

Regards,
Masoud Zargar[/quote]
use  a(4-a²)=0

4a=a^3 iff 4a-a^3=0 iff a(4-a^2)=0 iff a(2+a)(2-a)=0 so the solutions are: 0, 2 and -2 :P

marko


Given a pole of 20 feet and constant circumference of 3 feet.  There is a vine which winds seven times around the pole and reaches the top.  What is the length of the vine.
If you could please help in developing a formula to this question it would be most appreciated.

Ok, so we have a right triangle of height 20 feet, and as the vine winds around 7 times, the base is 7* 3pi. Just solve from there and you get that the length of the vine is about[hide]68.9[/hide] or so i think.

hmmm, this does not seem to work.  Any other idea's

azness has the right idea, just use 7*3 instead of 7*3pi
Unwrapping the vine from the pole would yield a right triangle of height 20 and base 21 (circumference times the # of times wrapped around).
Pythagorean Theorem yields
$\sqrt{21^2+20^2}=29$
So the vine is 29 feet long.


algorithm
number theory solved
number theory

Prove that there exists a sequence a_1<a_2<... of natural numbers such that a_n/(n^3) is bounded, but any positive integer can be uniquely expressed as a_i-a_j.

Use the greedy algorithme to construct such a sequence :

Let's choose $a_1=1,a_2=2$.
Now assume that $a_1 < a_2 < \cdots < a_k$ have been constructed so that $a_i < i^3 $, where $k > 1$, and such that all the differences $a_i-a_j$ are pairwise distincts.

Now, let $n$ be the smallest positive integer which is not of the form $a_i-a_j$. Then $n < k^2$ since there are exactly $\frac {k(k-1)} {2}$ such positive differences.
Let $a_{k+1}$ be the smallest positive integer such that neither $a_{k+1}$ nor $a_{k+1}+n$ are of the form $a_i-a_j+a_p$.
Note that it excludes less than $k^3$ integers (in fact, it excludes no more than $2C_k^3$ integers), thus $a_{k+1} < k^3$

If $a_{k+1} - a_i = n$ for some $i$, we are done. Otherwise, let $a_{k+2} = a_{k+1} + n < k^3 + k^2 < (k+1)^3$ and we are done.

Pierre.


greatest common divisor
number theory
prime factorization

:P Mr. Roland baked 252 cookies,  :) Mrs. Ashley baked 168 cookies, and  :lol: Ms. Nicole bakes 105 cookies. Each baker packaged them with the same number of cookies per package. What is the greatest number of cookies that could be in each package? (Round your answer to the nearest tenth if necessary.)

Why would you round to the nearest tenth for whole number amounts of cookies?

[quote="helga7032Z"][hide] the answer is 21 cookies per package.

[/hide] oops...you don't need to round to the nearest tenth. sorry. i 4got wut i was doing

 :rotfl:[/quote]
Uh...
I think you shouldn't give your answer before anyone even tries to solve the problem?
And anyways, it's just the GCF of the three numbers.

yes. it is simply the Greates COmmon Factor of teh three numbers but you still didn't answer what that is, ZhangPeijin. Do you know or not?

err, you just find the prime factorization of each number and find the lowest exponent for all of the common primes. then multiply. So yeah, its nothing extremely hard.

[quote="helga7032Z"]yes that is exactly how you do it. good job. you figured it out. (the reason it wasn't extremely hard is because it came from my mathbook) heehee. heehee.[/quote]
So I guess you are making us do your homework for you?
Or something else?
Besides, I know that if it were a learning mathbook, you don't need solutions, because they give you answer keys.


Prove that homtopic maps induce the same homomorphism on fundamental groups of  arc-wise connected sets up to an inner-automorphism.

I really hate how I've arranged this, as I know the result holds for $ \pi_n$'s as well (the factor you need to adjust for is the $ \pi_1$ action on $ \pi_n$), but this generalization is not immediate from the way I've decided to prove the result here. That being said, this was the first thing that popped into my head.

First of all, let's call $ x_0$ and $ y_0$ the basepoints of $ X$ and $ Y$, and $ f,g : X \to Y$ the homotopic maps. The result is only meaningful if $ f(x_0) \equal{} g(x_0) \equal{} y_0$, so let's assume that. Let $ H : X \times I \to Y$ be a homotopy from $ f$ to $ g$, such that $ H(x,0) \equal{} f(x)$ and $ H(x,1) \equal{} g(x)$. Let $ \gamma(t) \equal{} H(x_0,t)$. Let $ \alpha : I \to S^1 \to X$ be any loop.

Then consider $ K \equal{} H \circ (\alpha \times id) : I \times I \to Y$. We have that $ K(t,0) \equal{} f_*(\alpha)(t)$, and $ K(t,1) \equal{} g_*(\alpha)(t)$. Then this square $ K$ gives a homotopy from the loop $ f_*(\alpha)$ (running along the "bottom edge") to the loop $ \gamma g_*(\alpha) \gamma^{\minus{}1}$ (running the right edge, then the top edge, then the left edge in reverse). Therefore, in $ \pi_1(Y,y_0)$, we have that $ f_*(\alpha) \equal{} \gamma g_*(\alpha) \gamma^{\minus{}1}$. Note $ \gamma$ is independent from $ \alpha$, so $ f_*$ and $ g_*$ really only differ up to inner automorphism.


Find $a,b \in Z$ such that :

$C_{a+b}^{a}=2002$

Find all, or just one? Since you just said 'find', this is my answer.
$C_{2001+1}^{2001}=2002$

It's obviously ALL integers $a,b$ :roll:

nice solution. is everyone sure that it is absolutely fi9? to me its just all right. well done person.


number theory
relatively prime
number theory proposed

Find all odd positive integers $ n$ such that for any relatively prime divisors $ a$ and $ b$ of $ n$, the number $ a\plus{}b\minus{}1$ is also a divisor of $ n$.

[quote="Johan Gunardi"]Nope. You missed 25, 27, 121, etc.[/quote]

what!? i don´t think that you really are right.$ n$ can´t be 27.if you take $ a\equal{}3$ and $ b\equal{}9$ you´ll get a contradiction.i believe that you are wrong or the question is wrong. :huh:
by the way,if the question is indeed right ,the answer is indeed all primes ,just like it is already posted by modulamarc101. :wink:

[quote="Johan Gunardi"][quote="v235711"][quote="Johan Gunardi"]Nope. You missed 25, 27, 121, etc.[/quote]

what!? i don´t think that you really are right.$ n$ can´t be 27.if you take $ a \equal{} 3$ and $ b \equal{} 9$ you´ll get a contradiction.i believe that you are wrong or the question is wrong. :huh:
by the way,if the question is indeed right ,the answer is indeed all primes ,just like it is already posted by modulamarc101. :wink:[/quote]3 and 9 are not relatively prime. I am perfectly sure that the problem is correct.[/quote]

sorry i misread the problem. :blush:

[quote="Stephen"][quote="Rust"]Obviosly $ n \equal{} p^k$ is solution.[/quote]

But Why?[/quote]

That's because all of the factors of $ p^{k}$ is in the form $ p^{i}$ for some $ i\in \{0,1,2,...,k\}$.
Hence, the pair of relatively prime factors could be only $ 1,p^{i}$

By the way, if I didn't miss something, I just found that $ 12$ is a solution, too.
The problem is harder than I firstly thought!  :)

something along the lines of the following has to work:

Assume that $ n\equal{}2^k\cdot r$,where $ r$ is some odd $ r>1$ and $ k\geq 1$. So $ r\plus{}1$ divides $ 2^k\cdot r$. Thus, $ r\plus{}1\equal{}2^{l}$, where $ l$ is at least $ 2$, besides $ l\leq k$, so $ k\geq 2$,  therefore, either $ r\plus{}3$ divides $ 2^{k}$ or $ (r\plus{}3,r) \equal{} 3$.

The first means that $ (r\plus{}3) \minus{} (r\plus{}1)$ has to be a difference of two powers of two, but, it precisely means that the least of them is no more than $ 2$, it means $ l\equal{}1$.

The second leads to the conclusion, that $ 2^k \plus{} 3 \minus{} 1 \equal{} 2^k \plus{} 2$ divides $ 2^{k}\cdot r$, either, thus, $ 2^{k\minus{}1}\plus{}1$ divides $ r$, but as we know $ r\leq 2^{k} \minus{} 1$, so $ r \equal{} 2^{k} \minus{} 1$, therefore, $ 2^{k} \minus{} 1 \equal{} 2^{k\minus{}1} \plus{} 1$, so $ k\equal{}2$, and $ r\equal{}3$, hence $ \boxed{n \equal{} 12}$ or $ n\equal{} 2^{k}$.

Now let $ n$ be some odd integer, let $ q$ be its minimum divisor. And let $ p$ be the second prime divisor of $ n$ in the order after $ q$, i.e $ p> q$. Suppose $ n\equal{}p^{k}\cdot r$, where $ (r,p)\equal{}1$, clearly $ r\plus{}p\minus{}1$ divides $ n$.
Suppose that $ r\plus{}p\minus{}1$ and $ p$ are coprime, then $ r\plus{}p\minus{}1$ divides $ r$, what is impossible.

So $ r\plus{}p\minus{}1$ is divisible by $ p$, and so is $ r\minus{}1$. Clearly $ q$ is a prime divisor of $ r$, and let $ r\equal{}q\cdot t$, where $ t\geq q$, (we assume here that $ r$ is not prime).
Clearly, $ (t\plus{}p\minus{}1, p) \equal{} (t\minus{}1,p)$ and if $ p|t\minus{}1$, then $ p|q(t\minus{}1)$ and therefore $ p| q\minus{}1$, what is impossible. Hence, $ (t\plus{}p\minus{}1, p^k) \equal{} 1$

[b]Thus, $ t\plus{}p\minus{}1$ divides $ r$. [/b] Let $ (t\plus{}p\minus{}1,t) \equal{} m$, and let $ s$ be a prime divisor of $ m$, then $ s|t$ and $ s|p\minus{}1$, but then $ s$ has to be equal to $ q$, since $ s|n$. So $ m$ is a power of $ q$, and, that's not hard to deduce that $ t\plus{}p\minus{}1$ is a power of $ q$, either.

Doing almost the same process, for $ t\plus{}p^2\minus{}1$ and using that $ q>2$, we obtain a contradiction.

[quote="epsilonist"]By the way, if I didn't miss something, I just found that $ 12$ is a solution, too.
The problem is harder than I firstly thought!  :)[/quote]You missed something. Read the problem statement carefully, $ n$ is odd.

[hide="A hint"]I used the same notations as Erken. For any $ m$ divisor of $ r$, $ r<p\plus{}r\minus{}1<r\plus{}m$, thus $ p\plus{}r\minus{}1$ is relatively prime with $ r$. But we have $ p\plus{}r\minus{}1|n$. So $ p\plus{}r\minus{}1$ is a power of $ p$.[/hide]

Let $ p$ is minimal prime divisor of n. Because n - odd, p is odd too. Let $ a$ is another divisor of n, suth that $ (a,p)\equal{}1$. If there are not suth divisors, then $ n\equal{}p^k$. Were only $ a\equal{}1,,b\equal{}p^l$ satysfyed $ (a,b)\equal{}1$ and $ a\plus{}b\minus{}1\equal{}b$ divisor of n.
If $ n\not \equal{}p^k$, then $ n\equal{}p^kn_1, (n_1,p)\equal{}1$. If $ a\equal{}l\mod p$ divisor of $ n_1$, then  we had new divisors $ a\plus{}i(p\minus{}1), i\le l$ of n.
Therefore $ n_1\equal{}1\mod p$ (else $ n<\prod q_i, q_i|n, q_i,q_j)\equal{}1$) or $ n_1\equal{}p^l\minus{}p\plus{}1,l\le k$.
But for $ i\le k$ $ n_1\plus{}p^i\minus{}1\equal{}p^l\plus{}p^i\minus{}p\equal{}p(p^{l\minus{}1}\plus{}p^{i\minus{}1}\minus{}1)$ must be divisor of n. It give $ n<\prod q_i, q_i|n, (q_i,q_j)\equal{}1$.
Therrefore only $ n\equal{}p^k$ are solutions.


[quote="rd5493"]you want to find out?  ok, click here:

[url]http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=54[/url]

and try the chapter and then the state.  in my opinion, states was a pushover this year, but then again, i got a 27 at chapter, so anything better makes it seem like a pushover[/quote]

I got a 21 and chapter...  So states seemed like a school round.  (which we don't do)

[quote="egghead91"]I have never made it to states or countdown round... then again it is my first year. Can someone tell me (just in case I make it next year) if states are any different than chapter, only harder competition? :oops:[/quote]

The states competition is the exact same as chapter besides for in certain states they have a Master's Round where the top 4 have to come up with a presentation for a topic.  In WI we have the target round first at States, but I think that's just us.  The questions aren't really harder, it depends on the year.  For the countdown round, don't get nervous.  Other than that I have no more advice to give you.  Good luck next year!


limit
real analysis
real analysis unsolved

Problems like these have been bothering me for a while. For a quick example:
sqrt(20+sqrt(20+sqrt(20+sqrt(20... = 5
This can be written recursively as x(n+1) = sqrt.(x+x(sub.n))
And written explicitly as (1+sqrt.(1+4x))/2.

However I am not sure how to solve a problem such as the one following without using a calculator such as a TI-89 or repeatedly entering it into another calculator and observing the approached limit. On the test I found this problem calculators were not allowed so I assume there is an easier way:
sqrt.(11-6/sqrt.(11-6/sqrt.(11-6/sqrt.(11-6... = ?

Thank you for reading, and I apologize for the crude formatting, I am not sure how to include the proper symbols and such.

Here is a possible aproach:try to prove that the sequence is convergent and then take the recurence relation to limit:
For instance let's solve the following:

 $ x_{n}\equal{}\sqrt{20\plus{}\sqrt{20\plus{}..\plus{}\sqrt{20}}}$

Observer that : $ x_{n}<\sqrt{20\plus{}\sqrt{20\plus{}...\plus{}\sqrt{25}}}\equal{}5$
It's obvious that $ x_{n}$ is increasing.
From Weierstrass theorem $ \Rightarrow$ $ x_{n}$ converges.
Let $ \lim_{n\rightarrow \infty} x_{n}\equal{} \l$ $ \in \mathbb R_{\plus{}}$ .
Now taking the reccurence relation $ x_{n\plus{}1}\equal{}\sqrt{20\plus{}x_{n}}$ to limit we have :

$ \l\equal{}\sqrt{20\plus{}\l}$ ,wich gives $ l\equal{}5$ , since $ l>0$ , hence $ \lim_{n\rightarrow \infty} x_{n}\equal{}5$

Thank you. With your help I was able to solve my problem.


In colleges such as Stanford, MIT, Princeton and Duke
if you are interested in pursuing physics and economics double major (or somewhere along there), how many courses should you take each semester as a freshman?

Is it true that perhaps 3 is the best number of courses for the 1st semester, 1st year, and 4 for the 2nd semester?

Thank you  :jump:

Some colleges don't assign letter grades to first-year students so that you don't have to worry about this too much. Every college offers a choice of doing the easiest courses in the minimum amount, or taking more courses that are all harder. Wait till you are admitted to find out what the college's rules are, and what students admitted in earlier years advise about how to pursue a double major.

At top schools, classes are generally more intensive and challenging than at typical schools. It's probably not advisable to take six or more classes, unless you don't have any extracurricular activities at all. I took six classes one semester (Princeton) and it was rough. At Princeton they expect you to take about 4 classes per semester. You're supposed to do 5 once, and you only have to do 3 a semester when you're a senior. I personally find 5 to be the best courseload for me, but that's because I don't do much outside of classes.

At Stanford, 4 is pretty typical, though you'll often have an activity class (band, social dance, etc) thrown in.  It's reasonable to take 3 classes once or twice during your time, and I would strongly advise it fall quarter of freshman year.  Some people (like me this quarter) take 5 classes, but it can get difficult depending on what they are.

At MIT there's a credit limit for freshman.  First semester freshmen can only take up to 54 units.  So the majority of students either take 4 classes for 48 units or add on a seminar (or other 6 credit class) for 54 units.

Second semester freshmen can take 57 units which still only allows for 4 full classes.

That said, a nontrivial amount of students apply for sophomore standing their second semester here so they can take more classes.

I would suggest taking 4 classes both semesters freshman year.  It shouldn't be too hard especially if you have an easy schedule like me :D :(

[quote="joml88"]At MIT there's a credit limit for freshman.  First semester freshmen can only take up to 54 units.  So the majority of students either take 4 classes for 48 units or add on a seminar (or other 6 credit class) for 54 units.[/quote]

I should note for people familiar with other colleges that MIT counts units as putative hours that you will put into the course each week. At my undergraduate university, and at many others, what is counted is "credits." Usually, the rule of thumb is to multiply credits by 3 to get the number of hours per week that you will be busy

a) attending class,

b) working on labs,

and

c) doing homework

to keep up with the class. So 54 units in MIT-speak is like 18 credits at my alma mater (except that MIT's courses are probably harder than the courses at my alma mater anyway). A lot of colleges limit courses/credits/units for freshmen.

I'm an academic advisor at an institution rather different than those discussed previously in this thread, but some of the same principles exist anywhere. There are two forces in tension, that need to be balanced somehow.

1. You don't want to overload yourself into trouble. This is all going to be harder than your high school, and you won't have the 17,000 ways of recovering your grade from a bad test that you had in high school. In particular: any courses that you don't pass (including those you withdraw from) don't bring you any closer to graduation.

- but, on the other hand -

2. If you want to complete a "vertical" major like engineering or any science or mathematics, then you have to get it started. I use "vertical" to mean there exist long sequences of prerequisite courses. You can't take no science at all for a couple of years and then say, "Oh, by the way, I really wanted to major in EE." I'd certainly say that if you are a math/science/engineering major, then you should take at least one Math course every semester or quarter until you've taken all that you're going to.

Private schools tend to have different standards of what a "standard" courseload is.  Unless you are super-aggressive, don't take more than the standard courseload your first term.  After that, you can decide what is appropriate.  If you want to challenge yourself, do it by taking hard classes rather than lots of classes, but that's just my opinion.  Do start by taking one econ and one physics class your first term so you can decide if it's right for you.  Of course, this being aops, you'll want a math class, too, so you already have a pretty serious schedule lined up.


A chord $ ST$ of constant length slides around a semicircle with diameter $ AB$. Let $ M$ be the midpoint of $ ST$ and $ P$ the projection of $ S$ to $ AB$. Prove that $ \angle SPM$ is constant for all positions of $ ST$.

Let S' the symmetrical of the point S to the line AB
S' lie on the circle AB then $ \angle SS'T$ is constant (ST is constant)
therefore  $ \angle SPM$  is constant because $ \angle SS'T \equal{} \angle SPM$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Is there a nice solution for proving that there are exactly 9 types of finite subgroups of GL_2(Z)? I tried a lot to solve this problem, but I did not manage.

What about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33792]this problem[/url]? :? Maybe alekk meant $n>2$ there or something like that?

I do not see the relation with that problem. :?  I think he asks to prove that if G is a finite group then we can regard it as a subgroup of a certain $GL_n(Z)$ (the solution is not that difficult in this case: it suffices to regard G as a subgroup of permutations $S_n$ for a certain n and then to regard $S_n$ as a subgroup of $GL_n(Z)$ by associating to a permutation f the endomorphism which transforms $ (x_1,...,x_n)$ in $(x_{f(1)},...,x_{f(n)})$; by the way, there is a magnificient proof of Sylow theorem using this fact, due to Serre). As about $ GL_2(Z)$, I saw this in an article, but with no proof.

Oh, sorry. I thought alekk's problem asked for a proof that the statement is true for all $n$. :blush:

He couldn't have asked that, since in $GL_n(Z)$ any subgroup has at most $3^{n^2}$ elements.

Does it help if we know that $GL_2(\mathbb{Z})$ is generated by the matrices $\displaystyle \left(\begin{array}{cc} 1&0\\ 0&-1\end{array}\right)$, $\displaystyle \left(\begin{array}{cc} 0&1\\ -1&0\end{array}\right)$, and $\displaystyle \left(\begin{array}{cc} 0&-1\\ 1&1\end{array}\right)$?


$A$ is a set of $n$ non-negative integers. We say it has property $P$ if the set $\{x + y| x, y \in A\}$ has $\frac{n(n+1)}{2}$ elements. We call the largest element of $A$ minus the smallest element, the diameter of $A$. Let $f(n)$ be the smallest diameter of any set $A$ with property $P$. Show that $\frac{n^2}{4}\leq f(n) < n^3$.

[i]Brasil 1997[/i]

one year to pass, and no has one solution?  :maybe: