Without loss of generality, we can assume that $ a\geq c$, $ b\geq d$. Then
$ \frac{(a\minus{}b)(a\minus{}c)}{a\plus{}b\plus{}c}\plus{}\frac{(b\minus{}c)(b\minus{}d)}{b\plus{}c\plus{}d}\plus{}\frac{(c\minus{}d)(c\minus{}a)}{c\plus{}d\plus{}a}\plus{}\frac{(d\minus{}a)(d\minus{}b)}{d\plus{}a\plus{}b}$
$ \equal{}\frac{(a\minus{}c)^2}{a\plus{}c\plus{}d}\plus{}\frac{(b\minus{}d)^2}{a\plus{}b\plus{}d}\minus{}\frac{(a\minus{}b)(a\minus{}c)(b\minus{}d)(c^2\plus{}d^2\plus{}2ac\plus{}ad\plus{}bc\plus{}2bd}{(a\plus{}b\plus{}c)(a\plus{}b\plus{}d)(a\plus{}c\plus{}d)(b\plus{}c\plus{}d)}$.
If $ a\leq b$, then $ (a\minus{}b)(a\minus{}c)(b\minus{}d)\leq 0$, and if $ a\geq c$, $ b\geq c$, then $ (b\minus{}c)(b\minus{}d)(c\minus{}a)\leq 0$. Therefore, in these cases, we proved the inequality.
We only need prove the inequality when $ a\geq c\geq b\geq d$. Let $ x\equal{}a\minus{}c$, $ y\equal{}c\minus{}b$, $ z\equal{}b\minus{}d$. Then the inequality is equal to
$ \frac{x^2}{a\plus{}b\plus{}c}\plus{}\frac{z^2}{a\plus{}b\plus{}d}\plus{}xyz\left(\frac{1}{(a\plus{}c\plus{}d)(a\plus{}b\plus{}d)}\minus{}\frac{1}{(a\plus{}b\plus{}c)(a\plus{}c\plus{}d)}\minus{}\frac{1}{(a\plus{}b\plus{}d)(b\plus{}c\plus{}d)}\right)$ $ \geq 0$.
We know that
$ \frac{1}{a\plus{}b\plus{}c}\geq \frac{1}{b\plus{}c\plus{}d}\cdot \frac{y\plus{}2z}{x\plus{}2y\plus{}3z}\geq \frac{1}{b\plus{}c\plus{}d}\cdot \frac{y}{x\plus{}2y\plus{}3z}$.
Therefore,
$ \frac{x^2}{a\plus{}b\plus{}c}\plus{}\frac{z^2}{a\plus{}b\plus{}d}\geq \frac{x^2(x\plus{}y\plus{}2z)\plus{}z^2(x\plus{}2y\plus{}3z)}{(a\plus{}b\plus{}c)(a\plus{}b\plus{}d)}$
$ \geq\frac{(x^2\plus{}z^2)(x\plus{}2y\plus{}3z)}{2(a\plus{}b\plus{}c)(a\plus{}b\plus{}d)}\geq \frac{xyz}{(a\plus{}b\plus{}d)(b\plus{}c\plus{}d)}$.
Also, because $ (a\plus{}b\plus{}c)(a\plus{}c\plus{}d)\geq (a\plus{}c\plus{}d)(a\plus{}b\plus{}d)$,
$ xyz\left(\frac{1}{(a\plus{}c\plus{}d)(a\plus{}b\plus{}d)}\minus{}\frac{1}{(a\plus{}b\plus{}c)(a\plus{}c\plus{}d)}\right)\geq 0$.
Therefore, we proved the inequality when $ a\geq c\geq b\geq d$.
It is easy to see that the equality holds if and only if $ a\equal{}c$ and $ b\equal{}d$.

I will write just the main steps because I'm really tired since it's 3am

WLOG $ a \equal{} max(a,b,c,d)$

note that the inequality is equivalent to :

$ \frac {(a \minus{} d)(b \minus{} c)(a \plus{} 2d)}{(d \plus{} a \plus{} b)(c \plus{} a \plus{} d)} \plus{} \frac {(a \minus{} d)(b \minus{} c)(2b \plus{} c)}{(a \plus{} b \plus{} c)(b \plus{} c \plus{} d)} \plus{} \frac {(c \minus{} d)^2}{c \plus{} d \plus{} a} \plus{} \frac {(a \minus{} b)^2}{a \plus{} b \plus{} c}$

hence it remain to prove the inequality in the case $ c \geq b$

if $ a\geq d \geq c \geq b$
we have :
$ LHS \geq \frac {(a \minus{} b)(a \minus{} c)}{a \plus{} b \plus{} c} \plus{} \frac {(d \minus{} a)(d \minus{} b)}{d \plus{} a \plus{} b} \geq (d \minus{} b) ( \frac {a \minus{} c}{a \plus{} b \plus{} c} \plus{} \frac {d \minus{} a}{b \plus{} a \plus{} d} ) \geq \frac { (d \minus{} b)(d \minus{} c)}{b \plus{} a \plus{} d} \geq 0$

if  $ a\geq c \geq d \geq b$

  case 1 : $ a \minus{} c \leq d \minus{} b$

hence : $ a \minus{} d \leq c \minus{} b$
thus $ LHS \geq (a \minus{} c)( \frac {a \minus{} b}{a \plus{} b \plus{} c} \minus{} \frac { c \minus{} d}{c \plus{} d \plus{} a} ) \plus{} \frac { (a \minus{} c)(d \minus{} b)(c \minus{} b)}{(b \plus{} c \plus{} d)(d \plus{} a \plus{} b)} \geq 0$
case 2 :  $ a \minus{} c \geq d \minus{} b$

then the inequality is equivalent to :
$ (a \minus{} c)( \frac { a \minus{} b}{a \plus{} b \plus{} c} \minus{} \frac {c \minus{} d}{c \plus{} d \plus{} a} ) \plus{} (d \minus{} b)(\frac {d \minus{} a}{d \plus{} a \plus{} b} \plus{} \frac {c \minus{} b}{b \plus{} c \plus{} d} ) \geq 0$
thus  $ LHS \geq (d \minus{} b)( \frac { a \minus{} b}{a \plus{} b \plus{} c} \minus{} \frac {c \minus{} d}{c \plus{} d \plus{} a} \plus{} \frac {d \minus{} a}{d \plus{} a \plus{} b} \plus{} \frac {c \minus{} b}{b \plus{} c \plus{} d} ) \geq 0$  easy to prove !!

if  $ a\geq c \geq b \geq d$

note that the inequality is equivalent to :
$ (a \minus{} c)( \frac { a \minus{} b}{a \plus{} b \plus{} c} \minus{} \frac {c \minus{} d}{c \plus{} d \plus{} a} ) \plus{} (b \minus{} d) (\frac {a \minus{} d}{d \plus{} a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c \plus{} d} ) \geq 0$
now $ \frac {a \minus{} d}{d \plus{} a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c \plus{} d} \geq 0$  just let : $ a \equal{} d \plus{} x \plus{} y \plus{} z ,c \equal{} d \plus{} x \plus{} y....$

hence we have  2 cases :

if $ b \minus{} d \geq a \minus{} c$  as the previous method the inequality is easy to prove
if  $ a \minus{} c \geq b \minus{} d$
note that : $ \frac { a \minus{} b}{a \plus{} b \plus{} c} \minus{} \frac {c \minus{} d}{c \plus{} d \plus{} a} \geq 0$ is not true so we will use the trivial identity :
$ \frac { (a \minus{} c)(b \minus{} d)}{a \plus{} b \plus{} c} \plus{} \frac { (b \minus{} d)(c \minus{} a)}{a \plus{} b \plus{} c} \geq 0$

 hence the inequality is equivalent  to :
$ (a \minus{} c)( \frac { a \minus{} b}{a \plus{} b \plus{} c} \minus{} \frac {c \minus{} d}{c \plus{} d \plus{} a} \plus{} \frac {b \minus{} d}{a \plus{} b \plus{} c} ) \plus{} (b \minus{} d)(\frac {a \minus{} d}{d \plus{} a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c \plus{} d} \plus{} \frac {c \minus{} a}{a \plus{} b \plus{} c} ) \geq 0$
it is easy to prove that :$ \frac{ a\minus{}b}{a\plus{}b\plus{}c} \minus{}\frac{c\minus{}d}{c\plus{}d\plus{}a} \plus{} \frac{b\minus{}d}{a\plus{}b\plus{}c} \geq 0$with  the condition :  $ a \minus{} c \geq b \minus{} d$
thus : $ LHS \geq (b \minus{} d)( \frac {a \minus{} d}{b \plus{} c \plus{} a} \minus{} \frac {c \minus{} d}{c \plus{} d \plus{} a} \plus{} \frac {a \minus{} d}{d \plus{} a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c \plus{} d} \plus{} \frac {c \minus{} a}{a \plus{} b \plus{} c}) \geq 0$  the last inequality can be handeled with easy  manipulations , hence  the inequality is proved  :) omg what an ugly proof :(

[quote="April"]Prove that for any four positive real numbers $ a$, $ b$, $ c$, $ d$ the inequality
\[ \frac {(a \minus{} b)(a \minus{} c)}{a \plus{} b \plus{} c} \plus{} \frac {(b \minus{} c)(b \minus{} d)}{b \plus{} c \plus{} d} \plus{} \frac {(c \minus{} d)(c \minus{} a)}{c \plus{} d \plus{} a} \plus{} \frac {(d \minus{} a)(d \minus{} b)}{d \plus{} a \plus{} b}\ge 0\]
holds. Determine all cases of equality.

[i]Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany[/i][/quote]
Dear friend, please take a look at: http://mathvn.org/forum/viewthread.php?thread_id=1254 (problem O23, page 23).

My solution(very sorry if repeated):
Firstly it is equivalent to

$ \frac{(a-c)^2}{c+d+a}+\frac{(b-d)^2}{d+a+b}+(a-c)(b-d)(\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)}\ge 0$

We only need to prove:$ (\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)})^2\ge \frac{4}{(a+c+d)(b+d+a)}$

We are sufficient to show both $ (\frac{2b+d}{(a+b+d)(b+c+d)})^2$ and $ (\frac{2a+c}{(a+b+c)(c+d+a)})^2$ are not bigger than $ \frac{4}{(a+c+d)
(b+d+a)}$

They are almost trivial if we notice $ (2b+d)^2 \le 4(b+d)^2$ and some easy calculation :P

[quote="can_hang2007"][quote="April"]Prove that for any four positive real numbers $ a$, $ b$, $ c$,
[i]Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany[/i][/quote]
Dear friend, please take a look at: http://mathvn.org/forum/viewthread.php?thread_id=1254 (problem O23, page 23).[/quote]

Dear can_hang2007  I canot open  the file .pdf , please can you put here

My solution is more computational than the ones provided above, but in my opinion it is also a bit more motivated (though perhaps this could probably just be attributed to my lack of experience with inequalities.)

[hide]We shall show that if $a,b,c,d$ are any nonnegative real numbers such that $a \neq c$ or $b \neq d$

[b]Lemma 1[/b]: $x^4 - x^3 - x^2 + 2x + 1 \geq 0$ for all nonnegative $x$.
[i]Proof:[/i] We will use [url=http://en.wikipedia.org/wiki/Sturm%27s_theorem]Sturm's theorem.[/url]. Let $P_0(x) = x^4 - x^3 - x^2 + 2x + 1$. We compute the Sturm chain:
\begin{align*}
P_0(x) &= x^4 - x^3 - x^2 + 2x + 1\\
P_1(x) &= P_0'(x) &= 4x^3 - 3x^2 - 2x+2 \\
P_2(x) &= \frac{x}{4}(4x^3 - 3x^2 - 2x+2) - (x^4 - x^3 - x^2 + 2x + 1) &=& \frac{x^3}{4} + \frac{x^2}{2} - 2x + 1 \\
P_3(x) &= 16(\frac{x^3}{4} + \frac{x^2}{2} - 2x + 1) - (4x^3 - 3x^2 - 2x+2) &= 11x^2 - 30x + 14 \\
P_4(x) &= (\frac{x}{44}+\frac{13}{121})(11x^2 - 30x + 14) - (\frac{x^3}{4} + \frac{x^2}{2} - 2x + 1) &=\frac{-192x}{121} - \frac{61}{121} \\
P_5(x) &= (-\frac{192x}{121} - \frac{61}{121})(\frac{-1331x}{192} + \frac{778151}{36864}) - (11x^2 - 30x + 14) &= \frac{-908387}{36864}.
\end{align*}
(I could not get this to format properly; just right click on the image and click "view image.")

Note that $P_0(0) > 0$. By Sturm's theorem, the number of distinct roots in the interval $(0, \infty)$ is the difference in the number of sign changes of the leading coefficient of each polynomial in the Sturm chain and the number of sign changes in the constant term of each polynomial in the Sturm chain, which can easily be seen to be 0, so $P_0$ has no nonnegative real roots, as desired. $\blacksquare$

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(I shall rename $a,b,c,d$ as $w,x,y,z$, and call another set of variables $a,b,c,d$. This is due to my writing the latter portion, which is too difficult to change now, first.)

It is easy to see that the inequality is true when any three variables are equal, so we may suppose that there are not 3 equal variables among $w,x,y,z$. Without loss of generality, let $z$ be the smallest of $w,x,y,z$, and let $a = w-z$, $b = y-z$, and $c = w-z$. Define $f(k) = \frac{(a-b)(a-c)}{a+b+c+k} + \frac{(b-c)b}{b+c+k} + \frac{c(c-a)}{c+a+k} + \frac{ab}{a+b+k}$ (division by zero is not a problem here since no two of $a, b, c$ can equal 0, by our assumption that there are not 3 equal variables among $a,b,c,d$.) We wish to show that $f(z) \geq 0$.

The sign of $f$ is given by
$P(k) = (b+c+k)(a+b+c+k)(c+a+k)(a+b+k)f(k) \\
= (a^4 b + 2a^3 b^2 + 2a^2 b^3 + ab^4 + a^4 c - a^3 bc + 2ab^3 c + b^4 c - a^3 c^2 - 3a^2 bc^2 + 2b^3 c^2 - a^2 c^3 + 2abc^3 + 2b^2 c^3 + ac^4 + bc^4) + k(a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4) + k^2(2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3) + k^3((a-c)^2 + b^2).$
We claim that all coefficients of $P$ are nonnegative, which will finish this problem.

$a^4 b + 2a^3 b^2 + 2a^2 b^3 + ab^4 + a^4 c - a^3 bc + 2ab^3 c + b^4 c - a^3 c^2 - 3a^2 bc^2 + 2b^3 c^2 - a^2 c^3 + 2abc^3 + 2b^2 c^3 + ac^4 + bc^4 \geq 0$ is equivalent to $ac(a+c)(a-c)^2 + b(a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2) + b^2(2a^3 + 2a^2 b + ab^2 + 2abc + b^2 c + 2bc^2 + 2c^3).$ If $c = 0$, $a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2 \geq 0.$ Otherwise, substituting $w = \frac{a}{c}$ into lemma 1 gives $a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2 \geq 0$, so the constant coefficient of $P$ is nonnegative.

$a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4 \geq 0$ is equivalent to $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4ab + b^2 + 3bc + 4c^2) \geq 0.$ The inequality is obviously true if $c=0$, so suppose for now that $c \neq 0$. Note that it is sufficient to prove that $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4c^2) \geq 0.$ When viewed as a quadratic in $b$, the discriminant of the polynomial is $-a(a-c)^2(7a^3 + 60a^2 c + 52 ac^2 + 32c^3)$, which is clearly nonpositive. Hence, the coefficient of $k$ in $P$ is nonnegative.

$2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3 \geq 0$ is equivalent to $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a + 2b + 2c) \geq 0$, so it is sufficient to show that $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a+2c) \geq 0$. As before, we view the inequality as a quadratic in $b$. The discriminant of $b$ is $-(a-c)^2(20a^2 + 52ac + 7c^2) \leq 0$, so the coefficient of $k^2$ in $P$ is nonnegative.

Because the coefficient of $k^3$ in $P$ is clearly nonnegative, all coefficients of $P$ are nonnegative, as desired. It can easily be seen from the above inequalities that equality holds exactly when $a=c$ and $b=d$. [/hide]

math154 has pointed out that there is an error in this solution. $a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2 \geq 0$ is false (and is completely irrelevant to my lemma :(.) Here is a revised proof, hopefully with no mistakes this time:

[hide]We shall show that if $a,b,c,d$ are any nonnegative real numbers such that $a \neq c$ or $b \neq d$

[b]Lemma 1[/b]: If $x \geq 0$ and $(x-1)(x^3 - 3x - 1) < 0$, then $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1 \leq 0$.
[i]Proof:[/i]  Let $P(x) = x^3-3x-1$. $P(2) > 0$, $P(1) < 0$, $P(-1) > 0$, and $P(-2) < 0$, so by the intermediate value theorem, the three roots of $P$ lie in the intervals $(-2, -1)$, $(-1, 0)$, and $(1, 2)$. Since $P(0), P(1) < 0$, $P$ must be negative over $[0,1]$, so $(x-1)(P(x) \geq 0$ for $x \in [0,1]$. It follows that if $x \geq 0$ and $(x-1)(x^3 - 3x - 1) < 0$, we must have $x \in (1, 2)$.

We shall show that $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1 < 0$ for $x \in (1,2)$. The polynomial $3x^2 - 18x + 2$ has a minimum at $x=3$, so it is monotonic over $(1,2)$. Since it is negative at both $1$ and $2$, $3x^2 - 18x + 2 < 0$ for $x \in (1,2)$. Hence, $(1+4x)(3x^2 - 18x + 2) = 12x^3 - 69x^2 - 10x - 2 < 0$ for $x \in (1,2)$. Since $3x^4 - 23x^3 - 5x^2 + 2x - 1$ has derivative $12x^3 - 69x^2 - 10x - 2 < 0$ for $x \in (1,2)$ and is negative at $x=1$, $3x^4 - 23x^3 - 5x^2 + 2x - 1 < 0$ for $x \in (1,2)$. Hence, $2(1+x)(3x^4 - 23x^3 - 5x^2 + 2x - 1) = 6x^5 - 40x^4 - 56x^3 - 6x^2 + 2x - 2 < 0$ for $x \in (1,2)$. Because $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1$ is negative at $x=1$ and has derivative $6x^5 - 40x^4 - 56x^3 - 6x^2 + 2x - 2 < 0$ for $x \in (1,2)$, we have that $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1 < 0$ for all $x \in (1,2)$, as desired. $\blacksquare$

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(I shall rename $a,b,c,d$ as $w,x,y,z$, and call another set of variables $a,b,c,d$. This is due to my writing the latter portion, which is too difficult to change now, first.)

It is easy to see that the inequality is true when any three variables are equal, so we may suppose that there are not 3 equal variables among $w,x,y,z$. Without loss of generality, let $z$ be the smallest of $w,x,y,z$, and let $a = w-z$, $b = y-z$, and $c = w-z$. Define $f(k) = \frac{(a-b)(a-c)}{a+b+c+k} + \frac{(b-c)b}{b+c+k} + \frac{c(c-a)}{c+a+k} + \frac{ab}{a+b+k}$ (division by zero is not a problem here since no two of $a, b, c$ can equal 0, by our assumption that there are not 3 equal variables among $a,b,c,d$.) We wish to show that $f(z) \geq 0$.

The sign of $f$ is given by
$P(k) = (b+c+k)(a+b+c+k)(c+a+k)(a+b+k)f(k) \\
= (a^4 b + 2 a^3 b^2 + 2 a^2 b^3 + a b^4 + a^4 c - a^3 b c + 2 a b^3 c +
 b^4 c - a^3 c^2 - 3 a^2 b c^2 + 2 b^3 c^2 - a^2 c^3 + 2 a b c^3 +
 2 b^2 c^3 + a c^4 + b c^4) + k(a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4) + k^2(2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3) + k^3((a-c)^2 + b^2).$
We claim that all coefficients of $P$ are nonnegative, which will finish this problem.

$a^4 b + 2 a^3 b^2 + 2 a^2 b^3 + a b^4 + a^4 c - a^3 b c + 2 a b^3 c +
 b^4 c - a^3 c^2 - 3 a^2 b c^2 + 2 b^3 c^2 - a^2 c^3 + 2 a b c^3 +
 2 b^2 c^3 + a c^4 + b c^4 \geq 0$ is equivalent to $ac(a+c)(a-c)^2 + b(a-c)(a^3 - 3ac^2 + c^3) + b^2(2a^3 + 2a^2 b + ab^2 + 2abc + b^2 c + 2bc^2 + 2c^3) \geq 0$, so it is sufficient to show that $ac(a+c)(a-c)^2 + b(a-c)(a^3 - 3ac^2 + c^3) + b^2(2a^3 + 2c^3) \geq 0$. Since this inequality is homogeneous and obvious for $c=0$, we may assume without loss of generality that $c=1$,  so we wish to show that $a(a+1)(a-1)^2 + b(a-1)(a^3 - 3a - 1) + b^2(2a^3 + 2) \geq 0$. The result is obvious if $(a-1)(a^3-3a-1) \geq 0$, so suppose that $(a-1)(a^3 - 3a - 1) < 0$. Viewing this inequality as a quadratic in $b$, we see that it is sufficient to have $(a-1)^2(a^3 - 3a - 1)^2 - 4a(a+1)(2a^3 + 2)(a-1)^2 = (a-1)^2(a^6 - 8a^5 - 14a^4 - 2a^3 + a^2 - 2a + 1) \leq 0$, which is true by lemma 1. Hence, the constant coefficient of the polynomial is nonnegative.

$a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4 \geq 0$ is equivalent to $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4ab + b^2 + 3bc + 4c^2) \geq 0.$ Note that it is sufficient to prove that $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4c^2) \geq 0.$ We view the inequality as a quadratic in $b$. The discriminant of the quadratic is $-a(a-c)^2(7a^3 + 60a^2 c + 52 ac^2 + 32c^3)$, which is clearly nonpositive. Hence, the coefficient of $k$ in $P(k)$ is nonnegative.

$2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3 \geq 0$ is equivalent to $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a + 2b + 2c) \geq 0$, so it is sufficient to show that $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a+2c) \geq 0$. As before, we view the inequality as a quadratic in $b$. The discriminant of the quadratic is $-(a-c)^2(20a^2 + 52ac + 7c^2) \leq 0$, so the coefficient of $k^2$ in $P(k)$ is nonnegative.

Because the coefficient of $k^3$ in $P(k)$ is clearly nonnegative, all coefficients of $P$ are nonnegative, as desired. It can easily be seen from the above inequalities that equality holds exactly when $a=c$ and $b=d$. [/hide]

A7 in 2008 seems to be easy.
From $ a,b,c,d \in\mathbb{R}^+ $, we yield
 \[ \frac{(a-b)(a-c)}{a+b+c}\geq\frac{(a-b)(a-c)}{a+b+c+d}=\frac{a^2-ab-ac+bc}{a+b+c+d} \] Similarly, we got \[ \frac{(b-c)(b-d)}{b+c+d}\geq\frac{b^2-bc-bd+cd}{a+b+c+d} \] \[ \frac{(c-d)(c-a)}{c+d+a}\geq\frac{c^2-cd-ca+ad}{a+b+c+d} \] \[ \frac{(d-a)(d-b)}{d+a+b}\geq\frac{d^2-ad-bd+ab}{a+b+c+d} \] Sum all of the above inequalities together, we obtain: \[ \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b}\geq\frac{a^2-ab-ac+bc+b^2-bc-bc+cd+c^2-cd-ca+ad+d^2-da-db+ab}{a+b+c+d}=\frac{a^2+b^2+c^2+d^2}{a+b+c+d}\geq 0 \] Equality occurs iff $ a=c $ or $ b=d $

[quote=quangminhltv99]A7 in 2008 seems to be easy.
From $ a,b,c,d \in\mathbb{R}^+ $, we yield
 \[ \frac{(a-b)(a-c)}{a+b+c}\geq\frac{(a-b)(a-c)}{a+b+c+d}=\frac{a^2-ab-ac+bc}{a+b+c+d} \] [/quote]

Well, no. That holds only if the fractions are nonnegative. For example if $c>a>b$, then it doesn't hold.

[quote=hxy09]My solution(very sorry if repeated):
Firstly it is equivalent to

$ \frac{(a-c)^2}{c+d+a}+\frac{(b-d)^2}{d+a+b}+(a-c)(b-d)(\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)}\ge 0$

[/quote]

What would the motivation for splitting the expression in this way be? Thanks!

Just a query; after clearing the denominators, isn't it pretty much do-able by Lagrange Multipliers(of course, after normalizing)

well, hxy09...
[quote=hxy09]
We only need to prove$ (\frac{2b+d}{(a+b+d)(b+c+d)}-\frac{2a+c}{(a+b+c)(c+d+a)})^2\ge \frac{4}{(a+c+d)(b+d+a)}$[/quote]
But, this does not satisfy when $a=b=c=d$!!
.....isn't it!?  :wallbash_red:

[b]Note.[/b] The term ``LHS'' will always refer to the expression $\frac {(a - b)(a - c)}{a + b + c} + \frac {(b - c)(b - d)}{b + c + d} + \frac {(c - d)(c - a)}{c + d + a} + \frac {(d - a)(d - b)}{d + a + b}$.

We will rewrite the LHS as \[ (a-c)(\frac{a-b}{a+b+c} + \frac{d-c}{c+d+a}) + (b-d)(\frac{b-c}{b+c+d} + \frac{a-d}{d+a+b}). \] Let us focus on the first term $(a-c)(\frac{a-b}{a+b+c} + \frac{d-c}{c+d+a})$. We will try to express this in terms of $(a-c)$ and $(b-d)$. Thus we write \[(a-c)(\frac{a-b}{a+b+c} + \frac{d-c}{c+d+a}) = (a-c)(\frac{(ac+ad+a^2-bc-bd-ba) + (da+db+dc-ca-cb-c^2)}{(a+b+c)(c+d+a)}) \] \[ = (a-c)(\frac{2ad - 2bc - ab + cd + a^2 - c^2}{(a+b+c)(c+d+a)}) \] \[ = (a-c)(\frac{(a-c)(a+c+\frac{1}{2}b+\frac{1}{2}d) + (b-d)(-\frac{3}{2}a-\frac{3}{2}c)}{(a+b+c)(c+d+a)}) \] \[ = (a-c)^2\frac{a+c+\frac{1}{2}b+\frac{1}{2}d}{(a+b+c)(c+d+a)} + (a-c)(b-d)\frac{-\frac{3}{2}a-\frac{3}{2}c}{(a+b+c)(c+d+a)}\]

Similarly, the second term $(b-d)(\frac{b-c}{b+c+d} + \frac{a-d}{d+a+b})$ equals \[ (b-d)^2\frac{b+d+\frac{1}{2}a+\frac{1}{2}c}{(b+c+d)(d+a+b)} + (a-c)(b-d)\frac{\frac{3}{2}b+\frac{3}{2}d}{(b+c+d)(d+a+b)} \]

Thus, the entire LHS equals \[(a-c)^2X + (b-d)^2Z + (a-c)(b-d)Y,\] where $X = \frac{a+c+\frac{1}{2}b+\frac{1}{2}d}{(a+b+c)(c+d+a)}$, $Y = \frac{3}{2}(\frac{b+d}{(b+c+d)(d+a+b)} - \frac{a+c}{(a+b+c)(c+d+a)})$, and $Z = \frac{b+d+\frac{1}{2}a+\frac{1}{2}c}{(b+c+d)(d+a+b)}$. Let's simplify $Y$ a little bit. Miraculously, we can write \[Y = \frac{3}{2}(\frac{(b+d)(a+b+c)(c+d+a) - (a+c)(b+c+d)(d+a+b)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)})\] \[ = \frac{3}{2}(\frac{(b+d)(bd+(b+d)(a+c)+(a+c)^2) - (a+c)(ac+(a+c)(b+d) + (b+d)^2)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)})\] \[ = \frac{3}{2}(\frac{bd(b+d)-ac(a+c)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}).\] Note how most of the terms in the numerator cancel.

Let $u = a-c$, $v = b-d$. Then the LHS is \[u^2X + v^2Z + uvY,\] which is quadratic in $u,v$. I claim that $Y^2 - 4XZ < 0$, or equivalently:

[b]Lemma.[/b] $Y^2 < 4XZ$

[i]Proof.[/i] To prove this, we first note that \[4XZ = \frac{(2a+2c+b+d)(2b+2d+a+c)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\] and \[Y^2 = \frac{9}{4}(\frac{(bd(b+d)-ac(a+c))^2}{((a+b+c)(b+c+d)(c+d+a)(d+a+b))^2}).\] Thus $Y^2 < 4XZ$ is equivalent to \[\frac{9}{4}(bd(b+d)-ac(a+c))^2 < (2a+2c+b+d)(2b+2d+a+c)(a+b+c)(b+c+d)(c+d+a)(d+a+b)\]

Note that, if $x,y > 0$, $(x-y)^2 < x^2$ if $x \ge y$ and $(x-y)^2 < y^2$ if $y \ge x$. Let's let $x = ac(a+c)$, $y = bd(b+d)$. Then we can assume WLOG that $x \ge y$. We have \[\frac{9}{4}(bd(b+d)-ac(a+c))^2 < \frac{9}{4}(ac(a+c))^2\]

But we also have \[(2a+2c+b+d)(2b+2d+a+c)(a+b+c)(b+c+d)(c+d+a)(d+a+b) > (2a+2c)(a+c)(a+c)(c)(a+c)(a) = 2ac(a+c)^4\]

Thus, if we can show that \[2ac(a+c)^4 > \frac{9}{4}(ac(a+c))^2,\] we would be done with the proof of this Lemma. But this is equivalent to \[(a+c)^2 > \frac{9}{8}ac,\] which of course is true. $\blacksquare$

Now that we have $Y^2 - 4XZ < 0$, let's show that the LHS, which is $u^2X + v^2Z + uvY$, is always nonnegative. Note that $X,Z > 0$. First suppose that $u = v = 0$; then obviously the LHS equals $0$, which is nonnegative. Now suppose that $u = 0$ but $v \not= 0$. Then the LHS is $v^2Z$, which is positive. Similarly, if $u \not= 0 $ and $v = 0$, then the LHS is $u^2X$, which is positive.

Now suppose that $u,v \not= 0 $. We can write the LHS as \[v^2(X(u/v)^2 + Y(u/v) + Z)\] Since $Y^2 - 4XZ < 0$, we know that $Xr^2 + Yr + Z > 0$ for all real numbers $r$. This means that $X(u/v)^2 + Y(u/v) + Z > 0$. Since $v^2 > 0$, we know that the LHS is positive.

Thus the LHS is always nonnegative, with equality occurring if and only if $u = v = 0$, which is equivalent to $a=c$ and $b=d$ holding simultaneously. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	[The main rewrite] 	The given inequality may be rewritten as 	\begin{align*} 		(a-c)^2 & \left( \frac{1}{a+b+c} + \frac{1}{c+d+a} \right) 		+ (b-d)^2 \left( \frac{1}{b+c+d} + \frac{1}{d+a+b} \right) \\ 		&\ge 3(a-c)(b-d) 		\left[ \frac{a+c}{(a+b+c)(c+d+a)} 			- \frac{b+d}{(b+c+d)(d+a+b)} \right]. 	\end{align*}

[hide="Proof"]  Using the SOS identity $2(a-b)(a-c) = (a-b)^2 + (a-c)^2 - (b-c)^2$, we get \[ 2 \sum_{\text{cyc}} \frac{(a-b)(a-c)}{a+b+c} 	 = \sum_{\text{cyc}} \frac{(a-c)^2}{a+b+c} 	 + \sum_{\text{cyc}} \frac{(a-b)^2 - (b-c)^2}{a+b+c} \] We observe that the second sum should be divisible by $(a-c)(b-d)$. After \[ \sum_{\text{cyc}} \frac{(a-c)(a-2b+c)}{a+b+c} 	= \sum_{\text{cyc}} \left[ \frac{(a-c)(a-2b+c)}{a+b+c} + 2(a-c) \right] 	= \sum_{\text{cyc}} \frac{3(a-c)(a+c)}{a+b+c} \] we can pair up two of the terms via \begin{align*} 	\frac{3(a-c)(a+c)}{a+b+c} 	+ \frac{3(c-a)(c+a)}{c+d+a} 	&= 3(a-c)(a+c) \left[ \frac{1}{(a+c)+b} - \frac{1}{(a+c)+d} \right] \\ 	&= -3(a-c)(b-d) \cdot \frac{a+c}{(a+b+c)(c+d+a)}. \end{align*} Doing the same calculation on the other half finishes. $\blacksquare$[/hide]

In light of this, taking the absolute value of right-hand side, it suffices to prove \begin{align*} 	(a-c)^2 & \left( \frac{1}{a+b+c} + \frac{1}{c+d+a} \right) 	+ (b-d)^2 \left( \frac{1}{b+c+d} + \frac{1}{d+a+b} \right) \\ 	&\ge 3 |a-c| \cdot |b-d| \cdot 	\max \left\{ \frac{a+c}{(a+b+c)(c+d+a)}, 	\frac{b+d}{(b+c+d)(d+a+b)} \right\}. \end{align*} We show only the first one (other one ditto). In fact, a crude AM-GM suffices:

[b][color=red]Claim:[/color][/b]  	We have 	\[ 4\sqrt[4]{\prod_{\text{cyc}} \frac{1}{a+b+c}} 		> \frac{3(a+c)}{(a+b+c)(c+d+a)}.  \]

[hide="Proof"]It's equivalent to prove that 	\[ 256 (a+b+c)^3 (c+d+a)^3 > 81 (b+c+d) (d+a+b) (a+c)^4. \] 	We have $(b+c+d)(d+a+b) \le (b+d+\frac{a+c}{2})^2$ 	by Jensen on $x \mapsto \log(b+d+x)$. 	So if we let $x = \frac{b}{a+c}$ and $y = \frac{d}{a+c}$, 	then it is sufficient to prove 	\[ 256(x+1)^3 (y+1)^3 > 81(x+y+\tfrac12)^2. \] 	However $(x+1)(y+1) = xy + x + y + 1 > x+y+\tfrac12$ 	so this is obvious.[/hide]

And thus we are done. Equality holds iff $|a-c| = |b-d| = 0$, which is $a=c$ and $b=d$.


geometry
3D geometry
tetrahedron
Volume
geometric inequality
IMO
IMO 1967

Prove that a tetrahedron with just one edge length greater than $1$ has volume at most $ \frac{1}{8}.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[hide]Assume $ CD>1$ and let $ AB\equal{}x$. Let $ P,Q,R$ be the feet of perpendicular from $ C$ to $ AB$ and $ \triangle ABD$ and from $ D$ to $ AB$, respectively.

Suppose $ BP>PA$. We have that $ CP\equal{}\sqrt{CB^2\minus{}BT^2}\le\sqrt{1\minus{}\frac{x^2}4}$, $ CQ\le CP\le\sqrt{1\minus{}\frac{x^2}4}$. We also have $ DQ^2\le\sqrt{1\minus{}\frac{x^2}4}$. So the volume of the tetrahedron is $ \frac13\left(\frac12\cdot AB\cdot DR\right)CQ\le\frac{x}6\left(1\minus{}\frac{x^2}4\right)$.

We want to prove that this value is at most $ \frac18$, which is equivalent to $ (1\minus{}x)(3\minus{}x\minus{}x^2)\ge0$. This is true because $ 0<x\le 1$.[/hide]


modular arithmetic
number theory solved
number theory
Divisibility
Multiplicative NT
Multiplicative order

Suppose that $A$ be the set of all positive integer that can write in form $a^2+2b^2$ (where $a,b\in\mathbb {Z}$ and $b$ is not equal to $0$). Show that if $p$ be a prime number and $p^2\in A$ then $p\in A$.

[i]Marcel Tena[/i]

Let be $p^2=a^2+2b^2$ with $b\neq 0 \implies p \nmid b$. Then looking $\mod p$ gives that $z^2 = -2 \mod p$ has a solution, namely $z=ab^{-1}$.
Now take integers $x,y$ with $x \equiv zy \mod p$ and $|x|,|y| < \sqrt p, y \neq 0$ by Thue's lemma.
But then $x^2+2y^2 \equiv 0 \mod p$ and $0< x^2+2y^2<3p$.
If $x^2+2y^2=p$ we are done. In the other case $x^2+2y^2=2p$, $x$ is even, so $x=2x'$ and we get $y^2+2x'^2=p$.

Well in fact the more general requirment is true: Let $p$ is a prime. If $p$ is an divisor of an element in form $x^2+2y^2$ and $(p,y)=1$  then $p$ is in $A$.

Here's in fact the same with [b]ZetaX[/b] solution : assume that $a,b\in\mathbb Z$ which $b\neq0$ and $p\mid a^2+2b^2$ and $(p,b)=1$ we will find $c,d$ so that $p=c^2+2d^2$. If $p=2$ then chose $c=0$ and $d=1$. Assume that $p\geq3$. So we can find $x>0$ so that $x^2+2\equiv 0\pmod{p}$ (by mutiple the inverse of $d$ in modulo $p$.) Now consider all pair $(u,v)$ which $0\leq u,v\leq [\sqrt{p}]=k$ we have $(k+1)^2>p$ pair so have more than $p$ number in form $u+xv$ so we can find two distince pair $(u_i,v_i)$ which $i=1,2$ so that congurent to each other modulo $p$. Which gives us
\[ u_1-u_2\equiv x(v_2-x_1)\pmod{p}  \]
Chose $c=|u_1-u_2|$ and $d=|v_1-v_2$ then $c,d\neq0$ and
\[ c^2\equiv x^2d^2\equiv -2d^2\pmod{p}  \]
So $p\mid c^2+2d^2$. Now note that $0<2+2d^2<p+2p=3p$ so $c^2+2d^2=p$ or $2p$. If $c^2+2d^2=2p$ then $c$ is even and we chose $c_1=c/2, d_1=d/2$ we have $2c_1^2+d_1^2=p$. Of cause $c,c_1,d,d_1\neq0$ so we have $p$ must be longs to $A$.

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=470&t=150923 and http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=327

here there is realy an easier solution i think...
let p^2=a^2+2b^2

=> p^2-a^2=2b^2

=>(p-a)(p+a)=2b^2

and we know that p^2=1(mod 4)
 => b^2=0(mod 4) =>b=0(mod 2)
and (p-a, p+a)=2

=> p-a=2(x^2) and p+a=y^2 where x=1(mod 2) and y=0(mod 2)
or
p-a=x^2 and p+a=2(y^2) where x=0(mod 2) and y=1(mod2)

in each of these two conditions we have:
2p=2w^2+z^2 which w=1(mod 2)and z=0(mod 2)(=>z^2= (mod 4))

=>p=w^2+(z^2)/2 and we had z^2=0(mod 4)
so we could show that p is a member of A...

[hide=HINT] note that there exists infinite integers (a,b) with their gcd = p .. now according to Bezout Lemma : there exists integers (x,y) such that p=ax+by now choose x=az^2 and y=2q^2b and try showing that from this p^{2} exists in A


How many units apart is any pair of parallel sides of a regular hexagon with side of 6 units? Express your answer in simplest radical form.

[geogebra]d5f5a04b8ef85aef78fb20ba0d8c52a2aa8de571[/geogebra]

Notice that the triangle is a 30-60-90, which means it has the property that its hypotenuse is $ 2$ times the length of the shorter leg, while the longer leg is $ \sqrt{3}$ times the shorter leg. The length of the longer leg is thus $ \frac{6}{2}\sqrt{3}\equal{}3\sqrt{3}$. However, we want [i]twice[/i] the length of the longer leg, because that's the length between the two sides.
The answer is thus $ 3\sqrt{3}\times2\equal{}\boxed{6\sqrt{3}}$


Express $ \sin {4t} \plus{} \sqrt{3} \cos {4t}$ in the form $ R\sin {(4t\plus{} \alpha)}$. Thanks

For $ a\neq 0,\ b\neq 0,$ rewrite the expression in the form of

$ a\sin \theta \plus{}b\cos \theta \equal{}\sqrt{a^2\plus{}b^2}\left(\sin \theta \cdot \frac{a}{\sqrt{a^2\plus{}b^2}}\plus{}\cos \theta \cdot \frac{b}{\sqrt{a^2\plus{}b^2}}\right).$

[hide="Answer"]$ \sin 4t \plus{} \sqrt{3} \cos 4t \equal{} R \sin(4t \plus{} \alpha)$
$ \equal{}> \sin 4t \plus{} \sqrt{3} \cos 4t \equal{} R (\sin 4t \cos \alpha \plus{} \cos 4t \sin \alpha)$
$ \equal{}> \sin 4t \plus{} \sqrt{3} \cos 4t \equal{} (R \cos \alpha) \sin 4t \plus{} (R \sin \alpha) \cos 4t$

and so we must have

$ R \sin \alpha \equal{} \sqrt{3}$
$ R \cos \alpha \equal{} 1$.

Dividing both these equations we get $ \tan \alpha \equal{} \sqrt{3} \equal{}> \alpha \equal{} \arctan(\sqrt{3}) \equal{} \frac{\pi}{3}$. So, $ R \equal{} \frac{1}{\cos \alpha} \equal{} \frac{1}{\cos(\pi/3)} \equal{} 2$. Therefore,

$ \sin 4t \plus{} \sqrt{3} \cos 4t \equal{} 2 \sin\left(4t \plus{} \frac{\pi}{3}\right)$.[/hide]

@kunny, thanks... I'll be remembering that formula :D

@carcul, awesome solution, thanks :D


function
search
algebra proposed
algebra

Find all strictly increasing functions f from the naturals to the naturals such that:

f(f(n))=3n, for all naturals n.

Bomb

Posted before. Try to play around with some small values and try to generalise the things you notice. Look at where $3^k$ and $2 \cdot 3^k$ go. Or use the search button :p


By Cauchy-Schwartz in Engel Form,

$ \frac{a^6}{a^2\plus{}b\plus{}1}\plus{}\frac{b^6}{b^2\plus{}c\plus{}1}\plus{}\frac{c^6}{c^2\plus{}a\plus{}1}\ge \frac{(a^3\plus{}b^3\plus{}c^3)^3}{a^2\plus{}a\plus{}1\plus{}b^2\plus{}b\plus{}1\plus{}c^2\plus{}c\plus{}1}$

$ \equal{}\frac{9}{a^2\plus{}a\plus{}1\plus{}b^2\plus{}b\plus{}1\plus{}c^2\plus{}c\plus{}1}$

$ \equal{}\frac{3\plus{}\sum_{cyc}((\frac{2}{3}a^3\plus{}\frac{1}{3})\plus{}(\frac{1}{3}a^3\plus{}\frac{2}{3}))}{a^2\plus{}a\plus{}1\plus{}b^2\plus{}b\plus{}1\plus{}c^2\plus{}c\plus{}1}$

$ \ge\frac{3\plus{}\sum_{cyc}(a^2\plus{}a)}{a^2\plus{}a\plus{}1\plus{}b^2\plus{}b\plus{}1\plus{}c^2\plus{}c\plus{}1}\equal{}1$

My solution uses [b] Cauchy-Schwartz [/b] and the fact that: $ a^3\plus{}1 \ge a^2 \plus{} a$.


http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1292501#1292501

This is borderline on spam, but... :wink:

Problem:
Prove the AM-GM inequality. (No intervention from other states, please).

[quote="dragon96"]I from another state (sorry). Anyways, mewto was lying once again. AM-GM stands for Arithmetic mean-Geometric mean.[/quote]

 :rotfl:  :rotfl:  :rotfl:

He was pretending to be stupid...

Let X=The average MPG of the cars and let Y=the price of gas (in dollars)

BY the definitions of Asian and General Motors, $ AM \equal{} \frac {x \plus{} y}{2}$ and $ GM \equal{} \sqrt {xy}$

By the Automobile Theorem,

$ (x \minus{} y)^2\geq{0}$

$ x^2 \minus{} 2xy \plus{} y^2\geq{0}$

$ x^2 \plus{} 2xy \plus{} y^2\geq{4xy}$

$ (x \plus{} y)^2\geq{4xy}$

$ (x \plus{} y)\geq{2\sqrt {xy}}$

$ \frac {(x \plus{} y)}{2}\geq{\sqrt {xy}}$

$ AM\geq{GM}$

Equality holds true iff MPG=1 and Price of Gas= $ \$1$

Correct! Now, prove it in general. What if there are several more dimensions, such as z=the number of Asians allowed in the car?

I would use hybrid induction to prove for $ 2^k$ asians and then prove it for $ k\minus{}1$ asians.

[quote="tinytim"]I would use hybrid induction to prove for $ 2^k$ asians and then prove it for $ k \minus{} 1$ asians.[/quote]

Yup.
If it is true for $ a_1,a_2,\ldots,a_{n \plus{} 1}$, then we have

$ \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{n \plus{} 1}}{n \plus{} 1}\geq\sqrt [n \plus{} 1]{a_1a_2\cdots a_{n \plus{} 1}}$.

Substituting $ a_{n \plus{} 1} \equal{} \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}n$, we get

$ \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}n\geq\sqrt [n \plus{} 1]{a_1a_2\cdots a_{n \plus{} 1}}$

$ \frac {(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^{n \plus{} 1}}{n^{n \plus{} 1}}\geq a_1a_2\cdots a_n\cdot a_{n \plus{} 1}$

$ \frac {(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^n}{n^n}\geq a_1a_2\cdots a_n$

Taking the $ n$th root, we have

$ \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}n\geq\sqrt [n]{a_1a_2\cdots a_n}$.

Blah.

New Problem:
Prove the QM-AM-GM-HM Inequality. :wink:

Oh dang did this thread get cleaned out? MOD ALERT! (siren) MOD ALEERT!


geometry
trigonometry
calculus
geometry proposed

Let $ ABCD$ be a quadrilateral which has an incircle centered at $ O$. Prove that
\[ OA\cdot OC\plus{}OB\cdot OD\equal{}\sqrt{AB\cdot BC\cdot CD\cdot DA}\]

Beautiful identity!

I have just uploaded a note, "Circumscribed quadrilaterals revisited", on my website ( http://www.cip.ifi.lmu.de/~grinberg/ ), where this identity is Theorem 10 and a synthetic proof is given. Hope you will like it (although you will probably find most of the stuff quite well-known).

  Darij

[b]Very nice identity ![/b] Thanks,[i] Mekrazywong[/i] !
Indeed, this problem is totally trivial by trigonometry.
It is equivalently with the following conditioned identity:

$x+y+z+t=\pi\Longrightarrow\cos x\cdot \cos z +\cos y\cdot \cos t =\sin (x+y)\cdot \sin (y+z)\ \ (1)$.

Really it isn't worth proving syntetically, at the most
a syntetical prove for the conditioned identity (1) !

[u]Remark.[/u] In general,
$\cos x\cdot \cos z+\cos y\cdot \cos t =\sin (x+y)\cdot\sin (y+z)\Longleftrightarrow$$\cos \frac{x+y+z+t}{2}\cdot \cos y\cdot \cos \frac{x+y+z-t}{2}=0$.
[i]Mekrazywong[/i] and [i]Darij[/i], can give you a geometrical interpretation
(for a quadrilateral, without circumscribed) of this equivalence ?

First of all, I would like to clarify here this result was [i]not[/i] due to me. Actually this is a problem from Chinese TST 2003, proposed by a Chinese professor. By the way, there is a really simple solution via spiral similarity.

Denote by $ a\equal{}AB$, $ b\equal{}BC$, $ c\equal{}CD$, $ d\equal{}DA$ the lengths of the sides of $ ABCD$.
Let $ r$ be the ray of the inscribed circle. Then $ OA \equal{}\frac{r}{\sin\frac{A}{2}},OB \equal{}\frac{r}{\sin\frac{B}{2}},OC \equal{}\frac{r}{\sin\frac{C}{2}},OD \equal{}\frac{r}{\sin\frac{C}{2}}$.
Let $ M$ be the point where $ AB$ touches the circle. Then $ AM \equal{} rctg\frac{A}{2},\ MB \equal{} rctg\frac{B}{2}\Rightarrow a \equal{} AM\plus{}MB \equal{} r\cdot\frac{\sin\frac{A\plus{}B}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}$,
$ b \equal{} r\cdot\frac{\sin\frac{B\plus{}C}{2}}{\sin\frac{B}{2}\sin\frac{C}{2}},c \equal{} r\cdot\frac{\sin\frac{C\plus{}D}{2}}{\sin\frac{C}{2}\sin\frac{D}{2}},d \equal{} r\cdot\frac{\sin\frac{D\plus{}A}{2}}{\sin\frac{D}{2}\sin\frac{A}{2}}$.
Replacing $ OA,OB,OC,OD,a,b,c,d$, making some calculus and using the equalities $ \sin\frac{A\plus{}B}{2}\equal{}\sin\frac{C\plus{}D}{2}, sin\frac{B\plus{}C}{2}\equal{}\sin\frac{A\plus{}D}{2}$, the relation becomes
$ \sin\frac{A}{2}\sin\frac{C}{2}\plus{}\sin\frac{B}{2}\sin\frac{D}{2}\equal{}\sin\frac{A\plus{}B}{2}\sin\frac{B\plus{}C}{2}$.
But $ \sin\frac{A}{2}\sin\frac{C}{2}\plus{}\sin\frac{B}{2}\sin\frac{D}{2}\equal{}\frac{1}{2}\left[\cos\frac{A\minus{}C}{2}\minus{}\cos\frac{A\plus{}C}{2}\plus{}\cos\frac{B\minus{}D}{2}\minus{}\cos\frac{B\plus{}D}{2}\right] \equal{}$
$ \equal{}\frac{1}{2}\left(\cos\frac{A\minus{}C}{2}\plus{}\cos\frac{B\minus{}D}{2}\right) \equal{}\sin\frac{A\plus{}B}{2}\sin\frac{B\plus{}C}{2}$.

use this lemma :
we have that $ AB.BC \equal{} OB^{2} \plus{} \frac {OA.OB.OC}{OD}$

My solution:

Let $ E $ be a point satisfy $ \triangle CDE \sim \triangle BAO $ .

Since $ C, D, E, O $ are concyclic ,
so from Ptolemy theorem we get $ DE \cdot OC+CE\cdot OD=CD\cdot OE $ . ... $ (1) $

Since $ \frac{DE}{OA}=\frac{CE}{OB}=\frac{CD}{AB} $ ,

so from $ (1) $ we get $ OA\cdot OC+OB\cdot OD=AB\cdot OE $ . ... $ (2) $

Since $ \angle EOD=\angle ECD=\angle OBA=\angle CBO , \angle DEO=\angle DCO=\angle OCB $ ,

so we get $ \triangle DOE \sim \triangle OBC $ and $ \frac{OE}{BC}=\frac{OD}{OB} $ . ... $ (3) $

Similarly, we can prove $ \triangle OCE \sim \triangle AOD $ and $ \frac{OE}{DA}=\frac{CE}{OD} $ . ... $ (4) $

From $ (3), (4) $ we get $ \frac{OE^2}{BC\cdot DA}=\frac{CE}{OB}=\frac{CD}{AB} $ . ... $ (5) $

From $ (2), (5) $ we get $ (OA\cdot OC+OB\cdot OD)^2=AB^2\cdot \frac{BC\cdot CD\cdot DA}{AB}=AB\cdot BC\cdot CD\cdot DA $ .

Q.E.D

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Quadrilatere%20circonscriptible.pdf  p. 7...

Sincerely
Jean-Louis

Just solving this for no reason as I just came for this page for a different problem but saw this and wanted to gain some fake satisfaction before sleep of solving an [hide]easy[/hide] chinese tst.
Just consider the point of tangencies as $KEK'W$.Let $F$ be $KK' \cap EW$
Then consider a projective transformation that fixes  the incircle and sends $F$ to the center of the incircle.
Now ,we just have to prove the problem for a rhombus which is a mere Pythagorean .

[b]Solved with Max Lu. The work here is almost entirely copy-pasted from our work on USAMO 2004/6  :maybe: . [/b]

Let the incircle have radius $r$, and let the lengths of the tangents from $A,B,C,D$ be $a,b,c,d$. Then, note that
\[(w+ri)(x+ri)(y+ri)(z+ri) \text{is a negative real number}\]
since the argument of it is $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}+\frac{D}{2}=180$. Thus, we have $r\cdot \sum_{cyc} wxy - r^3 \sum_{cyc} a = 0\Longrightarrow r^2 = \frac{\sum_{cyc} wxy}{\sum_{cyc} w}$.

Now, note that this gives
\[a^2+r^2 = a^2 + \frac{\sum abc}{\sum a} = \frac{a^2\sum a + \sum abc}{\sum a} = \frac{(a+b)(a+c)(a+d)}{\sum a}\]

Using this, we get that
\[OA\cdot OC + OB\cdot OD = \sqrt{\frac{(a+b)(a+c)(a+d)}{a+b+c+d}} \cdot \sqrt{\frac{(a+c)(b+c)(c+d)}{a+b+c+d}}+ \sqrt{\frac{(a+b)(b+c)(b+d)}{a+b+c+d}} \cdot \sqrt{\frac{(a+d)(b+d)(c+d)}{a+b+c+d}} \]\[= \frac{((a+c)+(b+d)) \sqrt{(a+b)(a+d)(b+c)(c+d)}}{a+b+c+d} = \sqrt{AB\cdot BC\cdot CD\cdot DA}\]
and we're done!

[quote=rafaello]Let $X,Y,Z,W$ be touch points of the incircle of $ABCD$ and sides $DA,AB,BC,CA$, respectively. Let $A^*,B^*,C^*,D^*$ denote the inverses of $A,B,C,D$ of an inversion around $(XYZW)$, respectively. Note that $A^*,B^*,C^*,D^*$ are the midpoints of $XY,YZ,ZW,WX$, respectively.
[asy]
import olympiad;import geometry;
size(8cm);defaultpen(fontsize(10pt));

pair A,B,C,D,X,Y,Z,W,O,A0,B0,C0,D0;
X=dir(110);Y=dir(205);Z=dir(280);W=dir(10);O=(0,0);
A=intersectionpoint(perpendicular(X, line(X,O) ),perpendicular(Y, line(Y,O) ));
B=intersectionpoint(perpendicular(Y, line(Y,O) ),perpendicular(Z, line(Z,O) ));
C=intersectionpoint(perpendicular(Z, line(Z,O) ),perpendicular(W, line(W,O) ));
D=intersectionpoint(perpendicular(W, line(W,O) ),perpendicular(X, line(X,O) ));
A0=midpoint(X--Y);B0=midpoint(Y--Z);C0=midpoint(Z--W);D0=midpoint(W--X);

draw(circumcircle(X,Y,Z),royalblue);draw(A--B--C--D--cycle,heavyred+1);draw(A0--B0--C0--D0--cycle, fuchsia);draw(X--Y--Z--W--cycle, red+0.5);
draw(A--O--B,palered+1);draw(C--O--D,palered+1);

dot("$O$",O,dir(90));
dot("$X$",X,dir(X));
dot("$Y$",Y,dir(Y));
dot("$Z$",Z,dir(Z));
dot("$W$",W,dir(W));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$A^*$",A0,dir(A0));
dot("$B^*$",B0,dir(B0));
dot("$C^*$",C0,dir(C0));
dot("$D^*$",D0,dir(D0));
[/asy]
\begin{align*}
OA\cdot OC+OB\cdot OD&=\sqrt{AB\cdot BC\cdot CD\cdot DA}\Longleftrightarrow\\
\frac{r^2}{OA^*}\cdot\frac{r^2}{OC^*}+\frac{r^2}{OB^*}\cdot \frac{r^2}{OD^*}&= \sqrt{\frac{r^2\cdot A^*B^*}{OA^*\cdot OB^*}\cdot \frac{r^2\cdot B^*C^*}{OB^*\cdot OC^*}\cdot \frac{r^2\cdot C^*D^*}{OC^*\cdot OD^*}\cdot \frac{r^2\cdot D^*A^*}{OD^*\cdot OA^*}}\Longleftrightarrow\\
OA^*\cdot OC^*+OB^*\cdot OD^*&=\sqrt{A^*B^*\cdot B^*C^*\cdot C^*D^*\cdot D^*A^*}\Longleftrightarrow\\
OA^*\cdot OC^*+OB^*\cdot OD^*&=\frac{XZ\cdot YW}{4}\Longleftrightarrow\\
OA^*\cdot OC^*+OB^*\cdot OD^*&=\frac{XY\cdot ZW+YZ\cdot WX}{4}\Longleftrightarrow\\
OA^*\cdot OC^*+OB^*\cdot OD^*&=XA^*\cdot C^*Z+B^*Y\cdot D^*W\Longleftrightarrow\\
r^2(\sin{\angle YXO}\cdot \sin{\angle WZO}+\sin{\angle ZYO}\cdot \sin{\angle XWO})&=r^2(\cos{\angle YXO}\cdot \cos{\angle WZO}+\cos{\angle ZYO}\cdot \cos{\angle XWO})\Longleftrightarrow\\
\cos{(\angle YXO+\angle WZO)}&=-\cos{(\angle ZYO+\angle XWO)},
\end{align*}which is true as $\angle YXO+\angle WZO=\frac{2\pi-\angle XOY-\angle ZOW}{2}=\frac{\angle YOZ+\angle WOX}{2}=\pi-\angle ZYO-\angle XWO$. $\blacksquare$[/quote]

MOST BEAUTIFUL IDEE!(AND DIAGRAME!)


What is the force exerted by the atmosphere on one side of a two feet by three feet board? (Answer choices are given in lb)

the pressure times the area, so atmospheric pressure times 6..
silly imperial units!


geometry
algebra
polynomial
Vieta
area of a triangle
Heron\u0027s formula
algebra unsolved

The side lengths of a triangle are the roots of a cubic polynomial with rational coefficients. Prove that the altitudes of this triangle are roots of a polynomial of sixth degree with rational coefficients.

hayyyyyyy,,no one :? it's very beatiful ;)

ehsan2004

please!
could you send the solution??
This was also in greek IMO Tst 2005
(problem 1)

Suppose that $a,b,c$ are the roots of the polynomial
\[f(x)=x^3-mx^2+nx-p,
\]
with $m,n,p\in \mathbb{Q}$ and let $S$ denote the triangle's area.
Heron's formula and standard manipulations with Viete's formulas yield
\[S^2=-m^4+4m^2n-8mp,
\]
hence $S=\sqrt{A}$, for some $A\in \mathbb{Q}.$
We have $h_a=\frac{2S}{a}$, hence $a=\frac{2\sqrt{A}}{h_a}$. Write that $f(a)=0$, group the terms containing $\sqrt{A}$ and square again. After clearing the denominators, we find that $h_a$ is the root of a 6th degree rational polynomial.

I have a good solution.
Because the coefficients are rational we write the equation as
$ x^3-Ax^2+Bx-C=0 $
Now let a,b,c the lengths of the triangle.
a+b+c=2t=A
ab+bc+ca=B
abc=C
We get all these from Vieta.
And from Iron we have that $ E^2=t(t-a)(t-b)(t-c)=t(t^3+At^2+Bt-C)$ which is rational.
Also we have that
$\upsilon_a=\frac{2E}{a}$
$\upsilon_b=\frac{2E}{b}$
$\upsilon_c=\frac{2E}{c}$
are roots of $ (x^2-\frac{4E^2}{a^2})(x^2-\frac{4E^2}{b^2})(x^2-\frac{4E^2}{c^2})=0 $
This equation is 6th degree with rational coefficients.
There are rational because $ \frac{4E^2}{b^2}+\frac{4E^2}{c^2}+\frac{4E^2}{a^2}=\frac{4E^2}{C^2}(B^2-2AC) $ is rational
Also ${ \frac{16E^4}{b^2a^2}+\frac{16E^4}{b^2c^2}+\frac{16E^4}{c^2a^2}=
\frac{16(E^2)^2}{C^2}(A^2-2B)}$
The last is obviously rational so all are rational and we are done.

[hide="Solution"]
If $a, b, c$ are the sides, then the polynomial $(x-a)(x-b)(x-c)$ has rational coefficients.  Then the altitudes are $\dfrac{2\sqrt{s(s-a)(s-b)(s-c)}}{a, b, c}$.  By Vieta $s$ is rational, so $\sqrt{s(s-a)(s-b)(s-c)}$ is rational too.  For convenience, call this $q$.

Then $\Pi (x-\dfrac{2\sqrt q}{a})(x+\dfrac{2\sqrt q}}{a})=\Pi (x^2-\dfrac{4q}{a^2})$, and this clearly has rational coefficients.
[/hide]

This problem is not as daunting as it may seem.

Suppose $r_1, r_2, r_3$ are roots of some cubic $ax^3 + bx^2 + cx + d$. Then we have that by Vieta, $r_1 + r_2 + r_3 = \frac{-b}{a} \in \mathbb{Q}$, and $r_1r_2 + r_2r_3 + r_3r_1 = \frac{c}{a} \in \mathbb{Q}$, and finally $r_1r_2r_3 = -d/a \in \mathbb{Q}$. Hence each $r_i, 1 \leq i \leq 3$ is rational also.

Note that the altitudes are given by $2[\text{area}]/r_i$. Now the area is simply $\sqrt{s(s - r_1)(s - r_2)(s - r_3)}$, where $s$ is the semiperimeter (this is due to Heron's formula). Now for simplicity, let $X = \sqrt{s(s - r_1)(s - r_2)(s - r_3)}$. Consider the polynomial \[\left(x^2 - \frac{4X^2}{r_1^2}\right) \left(x^2 - \frac{4X^2}{r_2^2} \right) \left(x^2 - \frac{4X^2}{r_3^2} \right). \] Here each $\frac{4X^2}{r_i^2}$ is clearly rational. Hence this new polynomial will have rational coefficients, and also have degree $6$. $\blacksquare$


Determine all positive integers $n$ such that $n={d(n)}^2$.

Obviously so $ n$ is odd (a square has an odd number of divisors). Let $ n\equal{}p_{1}^{2k_{1}}\cdots p_{m}^{2k_{m}}$, then we have to solve \[ (2k_{1}\plus{}1)\cdots (2k_{m}\plus{}1)\equal{}p_{1}^{k_{1}}\cdots p_{m}^{k_{m}},\]
and since $ 2x\plus{}1<3^{x}$ for all $ x>1$ we have $ p_{1}\le 3$ and $ k_{1}\le 1$, while all other primes cannot appear. So $ n\in\{1,9\}$.

[hide=J12]Note that $n$ is a perfect square. So we know that $n=(p_1^{e_1}\cdot p_2^{e_2}...p_k^{e_k})^2$. We know that $\sqrt{n}=p_1^{e_1}\cdot p_2^{e_2}...p_k^{e_k}=d(n)=(2e_1+1)(2e_2+1)...(2e_k+1)$ so it follows that since $3^x>2x+1$ for $x>1$ that if $k_1>1$ we cannot have $p_1\ge3$. If $k_1=1$ then we have $p_1=3$ which is a solution. So $n=9$ is one solution. If $k_1>1$ and $p_1=2$ this won't work because we know that $d(n)$ is odd implying that $n$ is odd. The only case we need to check now is $n=1$ which is indeed a solution. So $n=1,9$ are our solutions.  [/hide]


linear algebra
matrix
MATHCOUNTS
floor function
function
vector
trigonometry

What does having numbers stacked together in tall parentheses denote? :blush: [/code]

I think he means combinations.

Look it up in the AoPS Wiki. It's \binom{top number}{bottom number}, in dollar signs, of course.

you mean like $\binom{a}{b}$?

it just means the number of ways a [i]b[/i] number of things can be chosen from an [i]a[/i] number of things.

its' just $\frac{a(a-1)(a-2)...(a-b+1)}{b!}$

Or more like $\binom{n}{r}=\frac{n!}{(n-r)!r!}$

$\lfloor$ $\rfloor$ What do those symbols mean?

It is the floor function/greatest integer function. $\lfloor n \rfloor$ denotes the greatest integer less than or equal to $n$.

For example:

$\lfloor 3.7 \rfloor = 3$
$\lfloor-12.32 \rfloor =-13.$

[url]http://en.wikipedia.org/wiki/Greatest_integer_function[/url]

I think you mean vectors.

Maybe you are allowed to just use basic matrix and vector stuff in MC forum(like addition, and that sort of stuff)

BTW i like pie, that i think means the determinant of that matrix

[hide]
since i am really dumb take the negation of every statement but this one and that is probably the truth[/hide]

[quote="rd5493"][quote="i_like_pie"]You mean like $\left|\begin{matrix}1&2\\3&4\end{matrix}\right|$?

Those are called matrices, and are above Mathcounts level.[/quote]

HOW ARE MATRICES ABOVE MATHCOUNTS LEVEL? I did matrices this year, and I'm in Algebra.  How matrces are too advanced for a forum that makes my head hurt day after day is beyond me.[/quote]

I think he means using matrices to transform vectors with trig and that kind of higher stuff.  Adding and subtacting them isn't above Mathcounts, buts it's useless for competitions.

Matrices dealing with linear algebra are above MC level. I'd be very angry if they asked a question about them.


Show (synthetically) that distance between vertices X and Y of two triangles
    XBC and YBC  is greater than distance between their  incenters.

    Lt.  Pestich

At least this has been discussed here: http://www.mathlinks.ro/Forum/viewtopic.php?t=15277 .

  darij


Find the coefficient of $ x^{3}y^{3}z^{2}$ in the expansion of $ (x\plus{}y\plus{}z)^{8}$.

I know you are supposed to use the Binomial Thereom but when I tried so i got the wrong answer.please  :help: me

[hide="Solution"]We can see that, before expanding, the coefficient of the variables are $ 1$, so we don't need to worry about that ($ 1^3$ and $ 1^2$ are all $ 1$, which changes nothing when multiplying).  First, we want to find out how many ways we can choose three $ x$ terms.  We can get that by doing $ \binom{8}{3}\equal{}56$.  Then, we choose our three $ y$ terms.  We can do that by $ \binom{8\minus{}3}{3}\equal{}\binom{5}{3}\equal{}10$.  We did $ 8\minus{}3$ because after we chose our $ x$ terms, we are left with only $ 5$ terms.  Therefore, our coefficient is $ 56\cdot10\equal{}\boxed{560}$[/hide]


I probably forgot a few... :lol:

I know, for many of these there isn't a lot of recordings out there to listen to (like Ivan Galamian and Pinchas Zukerman) but I thought it wouldn't do them justice not to include them.

Jascha Heifetz is the KING OF THEM ALL.  His technique is flawless and I feel so ashamed watching him play.   :wallbash_red:

Ivan Galamian was pretty good too.  He my teacher was one of his students.   :D

Actually, EVERYONE up there have made it into history books.

Yeah, I didn't vote for anyone because they're all beast.

Funny...Ivan Galamian was once my teacher's teacher too!

Scary... who's your teacher?  Are you seriously good at violin?   :huh:

I'll just say my teacher is staff at the North Carolina School of the Arts. And I'm not really that great at violin...not "seriously good" at least.

DUDE I'M GOING TO THE NORTH CAROLINA SCHOOL OF THE ARTS ON AUGUST 25TH.  :o

If his name is Kevin Lawrence (my teacher's friend and I think he's a student if Galamian too), then I've had a lesson with him and I think he's going to be my teacher when I go learn there.

This is extra extra extra scary... :o

Yay for Heifetz. According to my violin teacher, you need good ears to appreciate Heifetz' music. His sound is unique, and is recognizable from anywhere.

[quote="Quickster94"]Kevin Lawrence?  I think he was a faculty at the Green Mountain Chamber Music Festival[/quote]

Yeah I think so.  (He tried to get me to go to the Green Mountain Music camp thing but I was like, "nah"  :D )  In the school months though he works at the North Carolina School of the Arts.

oh, ok, my piano teacher was is a faculty at the festival, the only one actually from vermont i think  :maybe:

[quote="qwertythecucumber"]Whoops. Sorry.  :blush:

I wish to keep my violin teacher anonymous. It would probably do him/her shame to make it announced that I study under that person.  :rotfl:  :rotfl:[/quote]

Haha wow you're really self criticizing of yourself.

Ha for a second I almost forgot who Yehudi Menuhin was.

As I said thought, there's probably too many legendary violinists to list them all.   :lol:

Jascha Heifetz is my fave.

Isaac Stern is good too... Carnegie Hall is here only because of his generous support.   :lol:

dude, what? Leaving Oistrakh off a list of "favorite old violinists" is like leaving God off of a list of "most influential beings."

No. Not "unless you don't believe in God."

Omigosh! I can't believe I forgot Oistrakh!  :wallbash_red:  :wallbash_red:  :wallbash_red:

lol I don't actually know anything about violin, just sticking my atheist comments in where i can.

I don't believe that one person's atheistic views can change the influence of God on the rest of the world.   (but that's just me)

I don't either, but I just like to stick those kind of comments in to distinguish myself as an atheist :O

What about Milstein? But I agree, there's so many brilliant violinists it would be impossible to name all of them.

I like Milstein too!!! I definately think he should have been on this, but since he wasn't I voted for Heifetz.

If you're going to mention oistrakh then I think that you should probably include kogan. Obviously, both of these men were exceptional at what they did so its really hard to tell who's best. It's interesting to note how the soviet background affected both of these musicians. Overall, I believe that Kogan was overshadowed by Oistrakh due to the politics of the time.

Listen to this:

[youtube]xsFJS4I_05E[/youtube]

^ i never really liked his playing... he was very talented, but never one of my favorites.


[quote="Mark9182"]Solve for B given V = (1/3) bh[/quote]

[hide]
$V = \frac{1}{3}bh$

$\frac{V}{h}= \frac{1}{3}b$

$\frac{3V}{h}= b$
[/hide]

is that what you mean by solving for B?

Yes that is exactly what I thought the answer would be


During a company's first full year in business, the price of oil increased $ 50\%$. In the second year of business, the price of oil went down $ 50\%$ from its price at the beginning of that year. Its price at the end of the second year is what percent of its price at the beginning of the first year?

$ 1 \times 1.5 \times 0.5\equal{}0.75 \implies \boxed{75 \%}$.


I stumbled on this nice problem in AoPS Volume 1, and solved it.  I think it would be good on countdown.  Here it is:

$ \text{Find }x\text{ if }2^{16^x} \equal{} 16^{2^x}$

Can you solve it under 45 seconds?

It's really just a matter of setting $ 16\equal{}2^{4}$.

Then both bases are 2's so they are unimportant.


Three students play a card game. Assuming replacements are allowed, what is the probability the three students pick exactly the same card? What is the probability that all three students pick different cards? (Show work and necessary steps).

Well it depends on the number of cards, which you do not contain in the problem.
However, if you let n=# of cards, you would do it in the following way.

[hide="1 Solution"](1/n)^3[/hide]
[hide="2 Solution"]The first student could pick any card. The next student could pick any card but that card.  The last student could pick any card but those two.

So Probability = (n-1)(n-2)/(n 2 )[/hide]

[hide]If there are n unique cards, then it doesn't matter what the first student picks. There's a 1/n shot the second picks the same and 1/n shot the 3rd picks the same. So 1/n 2

Again, n unique cards. Doesn't matter what the first person picks. The second has a (n-1)/n shot. The third has a (n-2)/n shot. (n-1)(n-2)/n 2 [/hide]


algebra
polynomial
rotation
geometry
geometric transformation
function
complex analysis

Hello
Let $ A$ and $ B$ be polynomials with real coefficients such that all roots of $ A\plus{}iB$ have negative imaginary parts.
Show that for all $ u,v \in \mathbb R$ , all the roots of $ uA\plus{}vB$ are real.

This is better suited for the complex analysis subforum, so I'm moving it there.
[hide="Hint"]
Use argument principle in one of the half-planes to conclude that the curve $ t\mapsto (A(t), B(t))$, $ t\in\mathbb R$ rotates around the origin many times[/hide]

I don't know about that argument principle could you try to explain without it ? =/

Actualy i don't know anything about complex analysis that's why i posted it in real analysis
but how would the fact that the curve (A(t),B(t)) rotates a lot of time around the origin help proving the result ?

If a curve makes $ n$ full rotations around the origin, it has to cross each line through the origin $ 2n$ times. The number $ n$ here can be either an integer or a half-integer.

The argument principle is a very simple thing: it tells you that if the function doesn't vanish inside a closed curve $ \Gamma$, then the increment of the argument over $ \Gamma$ is $ 0$. In our case, we need to consider a contour $ \Gamma$ like this:
[asy]D((-3,0)--(3,0));
D((0,-1)--(0,3));
D(MP("-R",(-2,0))--MP("R",(2,0))--CR((0,0),2,0,180),red+linewidth(1));[/asy]
If the maximal degree of $ A$, $ B$, is $ n$, then the argument of $ A(z)\plus{}iB(z)$ increases by $ \pi n$ when $ z$ travels over half-circle (only the highest degree term matters there) whence it has to decrease by the same $ \pi n$ when you travel over $ [\minus{}R,R]$, which means that the curve rotates about the origin $ n/2$ times. You can also see it without the argument principle by factoring and looking at the behavior of each linear factor (it rotates by $ \pi$ clockwise) but I still wanted to bring up the argument principle anyway.

Of course, you need to be a bit careful since the last statement is not exact (you have a small error near endpoints), but it still proves the result for almost all $ (u,v)$ and then you can just use the fact that complex roots are stable.

I wonder if there exists some purely real-analytic (or, better, algebraic) solution.

Why does the argument of A(z) + i*B(z) increase by Pi*n when z travels over the half-circle ? :maybe:

Because, up to a small error, it is just the argument of $ cz^n$ there.

But then where do you use the fact that all roots have negative imaginary parts ?
EDIT :Okay i got it sorry, thanks  :D


[quote="Beat"]Prove that for $s,n\in N$ we have
$\sum_{r=1}^{s}\left(\frac{a_r}{a_r+a_{r+1}}\right)^n\ge\frac{s}{2^n}$.
Where $a_{s+1}=a_1$ and $a_r>0$
[hide]For $n=s=3$ we have Vietnamese TSE 05 [/hide][/quote]

Of course, It's not true.

The right problem has occured in my book.

Excuse, me.
The problem: Find all natural n and s for which this is true


inequalities
parameterization
inequalities proposed

Let $ {x,y,z}\in {R^{ \plus{} }},k\in{N},k\ge{2},$prove that
\[ (x^k \plus{} 1)(y^k \plus{} 1)(z^k \plus{} 1)\ge{\frac {3^{3\minus{} k}}{4}}(xy \plus{} yz \plus{} zx)^k,\]
and determine the condition of equation.

If you feel difficult, please just to prove the inequality for some specific values of parameter,which is holds.

[quote="fjwxcsl"]Let $ {x,y,z}\in {R^{ \plus{} }},k\in{N},k\ge{2},$prove that
\[ (x^k \plus{} 1)(y^k \plus{} 1)(z^k \plus{} 1)\ge{\frac {3^{k\minus{}1}}{4}}(xy \plus{} yz \plus{} zx)^k,\]
and determine the condition of equation.

If you feel difficult, please just to prove the inequality for some specific values of parameter,which is holds.[/quote]
I think you meant to prove
$ (x^k\plus{}1)(y^k\plus{}1)(z^k\plus{}1) \ge \frac{3^{3\minus{}k}}{4}(xy\plus{}yz\plus{}zx)^k,$ (*)
not
$ (x^k\plus{}1)(y^k\plus{}1)(z^k\plus{}1) \ge \frac{3^{k\minus{}1}}{4}(xy\plus{}yz\plus{}zx)^k.$ (**)
Because (*) holds, but (**) does not. For (*), we can prove it as follows

Since $ \frac{k}{2} \ge 1,$ for any positive real numbers $ a,b,c$, we have
$ \left( \frac{a^{\frac{k}{2}}\plus{}b^{\frac{k}{2}}\plus{}c^{\frac{k}{2}}}{3}\right)^{\frac{2}{k}} \ge \frac{a\plus{}b\plus{}c}{3},$
or
$ \left( {a^{\frac{k}{2}}\plus{}b^{\frac{k}{2}}\plus{}c^{\frac{k}{2}}}\right)^{2} \ge 3^{2\minus{}k}(a\plus{}b\plus{}c)^k.$
Setting $ a\equal{}xy,b\equal{}yz$ and $ c\equal{}zx,$ we get
$ \left( \sqrt{x^ky^k}\plus{}\sqrt{y^kz^k}\plus{}\sqrt{z^kx^k}\right)^2 \ge 3^{2\minus{}k}(xy\plus{}yz\plus{}zx)^k.$
According to this inequality, we see that it suffices to prove that
$ (u^2\plus{}1)(v^2\plus{}1)(w^2\plus{}1) \ge \frac{3}{4} (uv\plus{}vw\plus{}wu)^2,$
where $ u\equal{}\sqrt{x^k},v\equal{}\sqrt{y^k},w\equal{}\sqrt{z^k}.$

Now, using the Cauchy-Schwarz Inequality, we have
$ (uv\plus{}vw\plus{}wu)^2\equal{}[u(v\plus{}w)\plus{}vw]^2 \le (u^2\plus{}1)[(v\plus{}w)^2\plus{}v^2w^2].$
Therefore, it is enough to prove that
$ 4(v^2\plus{}1)(w^2\plus{}1) \ge 3[(v\plus{}w)^2\plus{}v^2w^2],$
or
$ (v\minus{}w)^2\plus{}(vw\minus{}2)^2 \ge 0,$
which is obviously true.

[quote="can_hang2007"][quote="fjwxcsl"]Let $ {x,y,z}\in {R^{ \plus{} }},k\in{N},k\ge{2},$prove that
\[ (x^k \plus{} 1)(y^k \plus{} 1)(z^k \plus{} 1)\ge{\frac {3^{k \minus{} 1}}{4}}(xy \plus{} yz \plus{} zx)^k,\]
and determine the condition of equation.

If you feel difficult, please just to prove the inequality for some specific values of parameter,which is holds.[/quote]
I think you meant to prove
$ (x^k \plus{} 1)(y^k \plus{} 1)(z^k \plus{} 1) \ge \frac {3^{3 \minus{} k}}{4}(xy \plus{} yz \plus{} zx)^k,$ (*)
not
$ (x^k \plus{} 1)(y^k \plus{} 1)(z^k \plus{} 1) \ge \frac {3^{k \minus{} 1}}{4}(xy \plus{} yz \plus{} zx)^k.$ (**)
Because (*) holds, but (**) does not. For (*), we can prove it as follows

Since $ \frac {k}{2} \ge 1,$ for any positive real numbers $ a,b,c$, we have
$ \left( \frac {a^{\frac {k}{2}} \plus{} b^{\frac {k}{2}} \plus{} c^{\frac {k}{2}}}{3}\right)^{\frac {2}{k}} \ge \frac {a \plus{} b \plus{} c}{3},$
or
$ \left( {a^{\frac {k}{2}} \plus{} b^{\frac {k}{2}} \plus{} c^{\frac {k}{2}}}\right)^{2} \ge 3^{2 \minus{} k}(a \plus{} b \plus{} c)^k.$
Setting $ a \equal{} xy,b \equal{} yz$ and $ c \equal{} zx,$ we get
$ \left( \sqrt {x^ky^k} \plus{} \sqrt {y^kz^k} \plus{} \sqrt {z^kx^k}\right)^2 \ge 3^{2 \minus{} k}(xy \plus{} yz \plus{} zx)^k.$
According to this inequality, we see that it suffices to prove that
$ (u^2 \plus{} 1)(v^2 \plus{} 1)(w^2 \plus{} 1) \ge \frac {3}{4} (uv \plus{} vw \plus{} wu)^2,$
where $ u \equal{} \sqrt {x^k},v \equal{} \sqrt {y^k},w \equal{} \sqrt {z^k}.$

Now, using the Cauchy-Schwarz Inequality, we have
$ (uv \plus{} vw \plus{} wu)^2 \equal{} [u(v \plus{} w) \plus{} vw]^2 \le (u^2 \plus{} 1)[(v \plus{} w)^2 \plus{} v^2w^2].$
Therefore, it is enough to prove that
$ 4(v^2 \plus{} 1)(w^2 \plus{} 1) \ge 3[(v \plus{} w)^2 \plus{} v^2w^2],$
or
$ (v \minus{} w)^2 \plus{} (vw \minus{} 2)^2 \ge 0,$
which is obviously true.[/quote]

Edited,thank you, you are right! Very nice proof,can_hang2007!


just wondering, how's the olympiads 2002 project going? I heard it was finished about half a year ago, still I haven't really heard of it since then. Or has the project stopped since the merge with AoPS?

Actually, as far as I know, the deadline was postponed to October 2004. BTW, for Valentin, Orlando, Pierre et al.: I am preparing another file with two solutions by Peter Scholze and one by me...

  darij

If you still need people to look for more solutions, I've got nothing better to do right now :)

I guess mathlinks is gonna publish a book with all the olympiads from 2002.  :)

To dig up an old cow: is this project still going? :)

I'm afraid not, (and I don't think this is the right place to discuss OP 02).


[quote="pkerichang"]I still haven't seen the message board for AIME seminar...

is it not up yet??[/quote]
I don't see it either, you're not the only one.

Yeah, I don't see it either. Where would it show up when it does?

I don't think there is one since class starts tommorow.
Maybe I'm mistaken though.

Mmm, I doubt it:
[quote="Email I got today from AoPS"]Also, there is a message board set up especially for this class.  Click on "Community" on the left side of the website, then scroll down to the "Online School" section of the Forum, in which you should see the "AIME Problem Seminar" message board.[/quote]


HI everybody!!! how did you all do your board exams??? :D

See the other thread abt CBSE Board exams or sth  :)
And why would Einstein be bothered abt the boards?  :P   :rotfl:

[quote="rituraj007"]u must have read al's biography(excerpt) previous year![/quote]

It was in 12th i believe  :D  And ofc their history is too fascinating to avoid a peek  :D

:oops: i was telling to albert einstein!!!!


Anyone doing it from the West Coast??? I am interested.

Rice and Stanford also hold very similar competitions, I believe.  You could always try doing one of them instead of making that huge commute.


Hi all,

Having some trouble solving this problem.

A group of 20 students, including 3 particular girls and 4 particular boys, are to be lined up in two rows with 10 students each. In how many ways can this be done if the 3 particular girls must be in the front row while the 4 particular boys be in the back?

The answer is (14C7)x(10!)^2. The (10!)^2 I have some understanding of, but I don't understand how the 14 came about in the binomial coefficient. The 7 I have more understanding of as well.

Thanks folks.

14 should be $ 20 \minus{} 4 \minus{} 3 \equal{} 13$ -- that term counts the number of ways to choose seven people to join the three selected girls in front.


algebra
polynomial
algorithm
abstract algebra
vector
Ring Theory
superior algebra

Consider the ideal generated by $<p^n,p^{n-1}x,p^{n-2}x^2...,px^{n-1},x^n>$ over $Z[x]$, where $p$ is a prime number. prove it is not generated by less than $n+1$ polynomials.

Can we characterize other ideals in $Z[x]$ whit the same property?

Let $I=<p,X>$. $I$ is a maximal ideal of $\mathbb{Z}[X]$. $I^n=<p^n,p^{n-1}X,\ldots,X^n>$.
The elements of $I^n$ have the form $\sum_{i=0}^np^{n-i}X^if_i(X)$ with $f_i\in\mathbb{Z}[X]\ \forall i=\overline{1,n}$.
Because $(X\mathbb{Z})\cdot I^n\subset I\cdot I^n=I^{n+1}$, it is easy to see that $\sum_{i=0}^np^{n-i}X^if_i(X)=\sum_{i=0}^np^{n-i}X^if_i(0) (mod\ I^{n+1})$. Therefore $I^n/I^{n+1}$ has a structure of $\mathbb{Z}$-module.
Because $(p\mathbb{Z})\cdot I^n\subset I\cdot I^n=I^{n+1}$, it follows that $(p\mathbb{Z})\cdot I^n/I^{n+1}=0$. Therefore $I^n/I^{n+1}$ is an $\mathbb{Z}_p$ vectorspace. It is obviously generated by $n+1$ elements, therefore $\dim_{\mathbb{Z}_p}I^n/I^{n+1}\leq n+1$.

Define $\varphi: \mathbb{Z}_p^{n+1}\to I^n/I^{n+1}$ by $\varphi(a_0,\ldots,a_n)=\sum_{i=0}^na_ip^{n-i}x^i$. ($x$ denotes the class of $X$ mod $I^{n+1}$).
We prove that $\varphi$ is injective. Let $\varphi(a_0,\ldots,a_n)=\sum_{i=0}^na_ip^{n-i}x^i=0$. Then $\sum_{i=0}^na_ip^{n-i}X^i\in I^{n+1}$ (this time we see the elements $(a_i)_{i=\overline{0,n}}$ as elements of $\mathbb{Z}$, not residues mod p).
So we have $\sum_{i=0}^na_ip^{n-i}X^i=\sum_{j=0}^{n+1}p^{n+1-j}X^jf_j(X)$ for some $f_0,\ldots,f_{n+1}\in\mathbb{Z}[X]$.
Reduce this mod p and obtain $a_nX^n=X^{n+1}f_{n+1}(X)$ in $\mathbb{Z}_p[X]$ which implies $a_n=f_{n+1}=0 (mod p)$.
This means that we have $\sum_{i=0}^{n-1}a_ip^{n-i}X^i\in I^{n+1}\Rightarrow \sum_{i=0}^{n-1}a_ip^{n-i-1}X^i\in I^n$ and we repeat this algorithm to prove $p|a_i\ \forall i=\overline{0,n}$. This proves that $\varphi$ is injective, so $dim_{\mathbb{Z}_p}I^n/I^{n+1}\geq n+1$.
Therefore $\dim_{\mathbb{Z}_p}I^n/I^{n+1}=n+1$.

Assume for a contradiction that $I^n$ is generated by $n$ elements. The images of these elements give a set of generatros for $I^n/I^{n+1}$ as $\mathbb{Z}$-module and $\mathbb{Z}_p$-module. This conflicts with the dimension of $I^n/I^{n+1}$ over $\mathbb{Z}_p$.

Yes, we find a similar Isomorphism, but it took me a lot to realise it worked, I mean, that the images work like a basis as a Vector space over $Z_p$. But it seems you did it to easy!!! We really had to work hard to prove the rest. Anyway there is a lot of theory I dont understand from your solution, do you ahve a good reference?

I don't think that what I did is easy. I don't quite yet understand my idea to consider $I^n/I^{n+1}$.
Good reference? I don't know. It is quite basic stuff in commutative ring theory like the fundamental isomorphism theorem, and you need some modules theory to uderstand the restriction/extension of scalars. I think you know most of this anyway if you got to where you said you did. Maybe some practice.
After all this I sure hope my solution is correct.

I think the idea of considering $I_n/I_{n+1}$ comes from the fact that we already ahd the clue of maping $I_n$ somehow to $Z_p^{n+1}$ So I first thought about some kind of projection, but i also was thinking about some inductive argument so I tried to relate $I_n$ and $I_{n+1}$ and I saw how only independent terms survived to my projection and later I found it was the action of seing the first ideal mod $I_{n+1}$, but later that multiples of $p$ where ruled out was a surprise, so I found a set i thought i could easily map to $Z_p^{n+1}$ but i didnt filled the details  so it wasnt working, but yesterday in class we finished it with some nice ideas but somehow more elementary than yours, we dont know so much theory (I mean , I).


What is an inverse function? I  couldn't do #13 since i didn't know what inverse function is. I am an 8th-grader so my knowledge in math is not great.

what was the question? to do inverse funcitons algebraically i usually just switch the y and the x and then solve for y...I don't know if the question you are thinking of is as simple as this.  but that's usually what i aim to do.
beans

wow i had no clue how to do this problem...spent too much time on it...

f(x) turned out to be -x-1, making the answer -2

maybe someone has a solution manual?

inverse functions are like f(f^-1(x)) = x, so like x/2 and 2x would be inverses of each other

the question was:

f(x) = ax+b, f^-1(x) = bx + a, solve for a+b

You can think of an inverse function, f-1(x), of a function f(x) as a function that "undoes" f(x). In other words, it's a function such that f-1(f(x)) = x.

For example, the inverse of f(x) = x2 is f-1(x) = :sqrt:x since f-1(f(x)) = :sqrt:(x2) = x.

The Problem:

f(x) = ax + b

f-1(x) = bx + a.

Find a+b.

Solution (in spoiler):

[hide]

x = f(f-1(x)) = a(bx+a) + b

              = abx + a2 + b

so the RHS is identically equal to the LHS, therefore

ab = 1,  and a^2 + b = 0.

=> b = -a2 = > -a3 = 1 = > a = -1.

=> -b = 1 = > b = -1.

therefore a + b = -1 + -1 = -2.

[/hide]

Geometrically, I can up with 1. I don't know where I messed up but I know it's wrong. :(

i plugged in 0 for x, to find f(0)=b.  Then f(b)^-1 = b^2 + a = 0, so b^2=-a.

Then i plugged in 1 for x, to get f(1)=a+b.  then f(a+b)^-1 = b(a+b) + a = 1.  Then substituted -b^2=a in that equation and solved.

"f^-1(f(x)) = x. " is the best way to do it

I did something weird that I think(hope) is correct.
The inverse function of f(x)=ax+b is equal to (x-b)/a
And we know (x-b)/a = bx+a

So 1/a=b and -b/a=a
So we can solve that a^3=-1 and then finding b=-1

Um...I found a=-1 and then I read the question as only asking for a so I just circled B and moved on! Gah!


Here is another conductor problem which I have difficulty with.  It is from Introduction to Electrodynamics by Griffith.

I am not cheating, because this is not homework for me.  It is just sth I'd like to understand.

Part b and d and e, I can basically guess.  But I really need a quantitative and qualitative understanding of why my answer might be wrong or correct.

Thanks so much!

a) is wrong. You drew the right Gaussian surface, but it's not true that $ Q_{enc}$ is just $ q_a$: you are forgetting about the induced charge. The reasoning is as follows:

At the surface of your Gaussian surface, the E-field is zero everywhere, since that is "in the meat of a conductor". Remember, within a conductor E is always zero! So we have $ \int \vec{E}\cdot d\vec{A} \equal{} \int 0\cdot d\vec{A} \equal{} 0$, which means $ Q_{enc} \equal{} 0$. But $ Q_{enc} \equal{} q_a \plus{} q_{induced}$, therefore the induced charge (on the surface of the sphere of radius a) equals $ \minus{} q_a$. Hence $ \sigma_a \equal{} \frac { \minus{} q_a}{4\pi\epsilon_0 r_a^2}$, and with the same arguments $ \sigma_b \equal{} \frac { \minus{} q_b}{4\pi\epsilon_0 r_b^2}$.

(the answers differ only in a minus sign, but the arguments are pretty different).

[b]Btw, you didn't compute $ \sigma_R$.[/b]
If you do that, I'll help you with the next questions (the depend in some way on your understanding of question a).

Very good, BlueVelvet.  Ops: before "some_other_one" notice the radius in the $ \sigma_b$ denominator should be $ r_b$ or better, b for this and a for the other to follow strictly Griffith (an unimportant glitch, but just in case...)

You're almost there, integration.

Dr. No should read this " 2.5.3 The Surface Charge on a Conductor; ..." In Griffith.

Thanks guys for bearing my stupidity  :D   It's always nice to get a check.  BTW, for part a, I basically forgot to put the minus, but I knew it, because Qe, as you said, had to be 0.

[quote="integration"]but I knew it, because Qe, as you said, had to be 0.[/quote]Ok, but why did you write $ \int \vec{E}\cdot d\vec{A}\equal{}\frac{q_a}{\epsilon_0}$? Because that's not correct ;)

b) and c) are correct.
d) is not: There can only be an electric force if there's an electric field. But in both cavities there is no E-field (except the one produced by the charges themselves of course). So the answer is just zero force on q_a and q_b.

e): you are right that $ \sigma_R$ will change, but what about the E-field outside?

As far as I recall , in such cases the charges can be assumed to be at the center and the field will just be like that of a uniformly charged sphere.But I cant understand the logic and dont know the proof either. :(

Ok, but why did you write ? Because that's not correct ...

That's just a bad habit I have, b/c whenever I see a charge inside a conducting cavity, I know it will cause a charge of opposite sign to be polarized on the inner surface.

d.  So basically the conductor is analogous to a faraday cage?  I once heard that you can turn on a crazy Van der graaff, charged with bunches of coulomb's next to you, but as long as you stay inside a conducting cage, you cannot "feel" the outside electric field, because charge polarization on the cage causes it to cancel the outer electric field, on the inside.

So you're correct about no force, but do you have a kind of intuitive explanation?

e.  Well, if sigma r changes, the field outside will be the superposition of the answer in (b), with a different sigma r, and the field of the outer charge.  Is that correct?

Thx.

[quote="integration"]d.  So basically the conductor is analogous to a faraday cage?  I once heard that you can turn on a crazy Van der graaff, charged with bunches of coulomb's next to you, but as long as you stay inside a conducting cage, you cannot "feel" the outside electric field, because charge polarization on the cage causes it to cancel the outer electric field, on the inside.

So you're correct about no force, but do you have a kind of intuitive explanation?[/quote]
Well, I think yours is pretty intuitive: as long as you are inside a conducting cage (a [i]perfect [/i]conductor, and of course you are not touching it), the conductor makes sure there will be no E-field inside it. It certainly is comparable with a Faraday cage, in fact this [i]is[/i] a Faraday cage.
[quote]e.  Well, if sigma r changes, the field outside will be the superposition of the answer in (b), with a different sigma r, and the field of the outer charge.  Is that correct?[/quote]Yep, that's what I was thinking too.


how would you type a direct sum symbol in mathmode?

$ \sum\frac{1}{a\plus{}b}$

\sum\frac1{a+b}

$ \sum_{\text{sym}}abc$

\sum_{\text{sym}}abc

$ \sum_{k\equal{}0}^{\infty}\frac{x^{k}}{k!}$

\sum_{k=0}^\infty\frac{x^k}{k!}

Did you mean something like that?

the direct summand, it's an encircled plus sign. look at the link:

http://www.sosmath.com/CBB/latexrender/pictures/85f87586990c2b3c10d850c862f148f2.gif


What is everyone's state cutoff score? MA is 37.

KY: 32,33, or 34 (i'm sure its one of those scores)

We've already got one of these, I'll lock this because the other has more posts ATM.


[quote="rt_08"]not a GOOD SIGNATURE[/quote]

... that was random.

The USAMTS entry form specifies that you should send the form in separately through mail or fax (no email!) to the USAMTS people by the round 1 due date (10/14/08), so I don't think you have to send the entry form first as long as it gets there on time.

Why would you want to send in the solutions first? Have you already submitted them before you sent in the entry form?

Thanks.

Yeah I finished my solutions and I want to send them in, I was just wondering if I should wait until after my entry form goes in.

[quote="buzzer11"]I am submitting my solutions online, do I have to submit my entry form by snail mail?
Is it okay to submit my solutions, then my entry form, as long as both come in on time?[/quote]
I think it's more appropriate to submit your entry form first, then your solutions. Entry forms can be submitted in any way, mail or fax. As long as USAMTS receives it with the proper signatures, then you're fine. :)

buzzer's sig is sick i dont know what rt_08 is talking about.....

Haha, thanks mathemonster and thanks everyone for the replies...

I faxed in my entry form today and then my solutions tonight so I think i should be okay!


inequalities
geometry
inradius
inequalities unsolved

I have just found a good solution using Jensen's inequality. But I don't like it because I must prove Jensen again when using it. Please help me find another way.

[quote="darkmaster"]Prove that in a triangel ABC :
  (tan(A/2))^(2 \sqrt 2) + (tan(B/2))^(2 \sqrt 2) + (tan(C/2))^(2 \sqrt  2) > = 3^(1- \sqrt 2)[/quote]

Well, let t = 2 \sqrt 2. Then, your inequality can be rewritten as

  (tan(A/2))^t + (tan(B/2))^t + (tan(C/2))^t > = 3^(1 - t/2).

I will prove this inequality for all t >= 1.

At first, by the power mean inequality,

  ( ( (tan(A/2))^t + (tan(B/2))^t + (tan(C/2))^t ) / 3 ) ^ (1/t) >= ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3.

In other words,

  (tan(A/2))^t + (tan(B/2))^t + (tan(C/2))^t >= 3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t.

If we are now able to show that

  3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 3^(1 - t/2),

then we are done. Perform some equivalence transformations with this inequality:

  3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 3^(1 - t/2)
<==>  3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 3 / (sqrt(3))^t
<==>  ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 1 / (sqrt(3))^t
<==>  ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 >= 1 / sqrt(3)
<==>  tan(A/2) + tan(B/2) + tan(C/2) >= sqrt(3).

Now, this inequality is well-known and can be proven without Jensen. For instance, call s = (a + b + c) / 2 the semiperimeter and r the inradius of triangle ABC, then the inequality becomes

  r / (s-a) + r / (s-b) + r / (s-c) >= sqrt(3),

or

  1 / (s-a) + 1 / (s-b) + 1 / (s-c) >= sqrt(3) / r.

This was proven on page 5 of [url=http://www.mathlinks.ro/download.php?id=115]my MathLinks Contest 2 PDF file[/url].

  Darij


ACE angle is 180 degrees
Let there be  $T$ the middle of $BC$.
The point where $AB$  intersects $ET$ is ${S}$.
Prove that $SB=\frac{CD}{3}$

[quote="alex__ib"]$\angle ACE = 180^\circ$
Let $T$ be the midpoint of $BC$.
$S$ is the intersection of $AB$ and $ET$.
Prove that $SB=\frac{CD}{3}$[/quote]

Let's assume $B$ is not on $AE$ and $AB$ is not parallel to $ET$.

What is $CD$?? :maybe:

let there be ABC and DCE two equilateral congruent triangles
look at the description thingy or the title :wink:

[quote="vishalarul"]It's possible to make a diagram where segment $AB$ doesn't intersect segment $ET$.  :wink:[/quote]
He probably means line $AB$ and line $ET$
Those always intersect  :wink:


just wondering what you guys are going to be doing...whether practicing more maths/other subjects, cooling off, preparing for IMO/other olympiads, just to know what you guys are planning to do...  :D

let me start off, i'll be going to the beach this weekend and then with my cousins we are going to Six Flags!!! this is my first time so i should enjoy it  :play_ball:  :icecream:  :trampoline:

[quote="furious"]i'll probably be staying home doing the usual. probably get some fireworks as well. nothin' special.[/quote]

So will I  :D  No Rhode Island for me this year.  :(

my first 4th of july in NC will be my first firework setting 4th of july ever. (Cali doesn't let you buy fireworks; what a bummer :( )

i'm gonna go get ready to go to camp at duke so i have to buy stuff to bring there and study

I wonder what will happen for 4th july when I go to HCSSiM....hmmmm...

I am probably coming back from a wedding to which I am going to tomorrow!

[quote="math92"]I'm staying home or watching War of the Worlds[/quote]
im sorry but i think that was an overrated movie, i didnt like the ending! too cheesy
the movie i recommend? [color=blue]Batman Begins[/color]

[quote="Palytoxin"]just wondering what you guys are going to be doing...whether practicing more maths/other subjects, cooling off, preparing for IMO/other olympiads, just to know what you guys are planning to do...  :D [/quote]I'll be flying across the big water to check out the Fireworks in SD :)

I'm doing math!!! :D  :lol:  :lol:  :D
I'm also going on vacation, but I don't know where yet :(

War...of...the worlds OVERRATED it is one of the better movies out there, and if you really want to know about the "cheasy" ending read the book. If anything is overrated its Batman.

Edit: Ok for the weekend pick War of the Worlds its a much more enjoyable movie and I reccommend it more than Batman

P.S. Read the book first

[quote="sirkuku"]War...of...the worlds OVERRATED it is one of the better movies out there, and if you really want to know about the "cheasy" ending read the book. If anything is overrated its Batman.[/quote]I think you got the topic wrong ...  :?

[quote="sirkuku"]War...of...the worlds OVERRATED it is one of the better movies out there, and if you really want to know about the "cheasy" ending read the book. If anything is overrated its Batman.[/quote]
we are talking about what we are doing over the weekend not what movie is overated.
My sister, colts18, went to some resort in pensylvania called the Nemacolin.  She rode in a hummer, played paintball guns, rode in a fourth of july float, and did rock climbing.


calculus
derivative
function
calculus computations

The rate of change of a population depends on the current population $ P$ and is given by $ \frac{d}{dt}\equal{}kP(L\minus{}P)$ with $ K$ and $ L$ being positive constants.

How do you find for the nonnegative values of $ P$ is where population is increasing, decreasing, and constant?

The double derivative of the function?

The current population is given by $ P$, so the rate of change is the derivative of $ P$ with respect to time (as mentioned). When $ P$ is increasing, decreasing, or constant, the first derivative will be positive, negative, or zero (respectively), because it represents instantaneous rate of change.

Introducing second derivatives is asking for rate of rate of change, which isn't what you want.

So I guess the issue I have is how do you find the derivative and second derivative without $ t$ being a part of the equation you are taking the derivative of.


Figure $ABCD$ is a trapezoid with $AB || DC, AB = 5, BC = 3 \sqrt 2, \measuredangle BCD = 45^\circ$, and $\measuredangle CDA = 60^\circ$. The length of $DC$ is

$\textbf{(A) }7 + \frac{2}{3} \sqrt{3}\qquad
\textbf{(B) }8\qquad
\textbf{(C) }9 \frac{1}{2}\qquad
\textbf{(D) }8 + \sqrt 3\qquad
\textbf{(E) }8 + 3 \sqrt 3$

[hide="Answer"]Because we see the angle measures 45 and 60, we look for special right triangles. By splitting the trapezoid into a 30,60,90 triangle, 445,45,90 triangle, and a rectangle, we can find CD piece by piece.

$CD=\sqrt{3}+5+3=8+\sqrt{3}$
The answer is D.
[/hide]

[hide="Answer"]We have $BC=3\sqrt{2}$, and $\angle BCD=45^\circ$, so, labeling the altitude to $B$ from $DC$ as $BF$, we have $BF=3$ and $CF=3$. Labeling the altitude from $A$ to $DC$ as $AE$, we have $AE=3$, and since $\angle CDA=60^\circ$, then $DE=\sqrt{3}$, so we have $5+3+\sqrt{3}=8+\sqrt{3}\Rightarrow \boxed{D}$.[/hide]

[hide]$DC=AB+x+y$, well looking at the triangles you are given, which are $45,45,90$ and $30,60,90$, you can see that the height of the trapezoid is $3$, therefore $x=3$ and $y=\sqrt{3}$, therefore $\boxed{DC=8+\sqrt{3}}$[/hide]


same questions daniel had. it seems impossible for me to include any diagram in my answers.

If you are going to add your own points or labels, then you should definitely include a diagram. You can read the $\LaTeX$ tutorial on the AoPS website for information on how to include diagrams in your $\LaTeX$ files.

[quote="bookworm271828"]Just wondering, do our answer to geometry questions need to include the diagram, or can we just use not given labels and not worry about including it.

Thanks for your help.[/quote]

You don't need to have a diagram, but it sure does help if it's at all different from the problem.

Regarding diagrams, if you're too lazy to use the xy-pic environment and whatnot to generate diagrams (like me), then you can simply make them in paint, save them as JPG files in the same folder as your LaTeX document, and use the command \includegraphics{filename} in your LaTeX document to embed them.  It's not the most compact or the prettiest form, but it's by far the simplest.

Still, I recommend reading the picture section in the LaTeX tutorials.  That's what it's there for!


Suppose a function $f:$ $N \rightarrow N$, $ff(n) = kn , k \in R$
Prove that
$\frac{2k}{k+1}n \leq f(n) \leq \frac{k+1}{2}n$
And find an example of $f(n)$ when $k = 3$

[quote="Kalimdor"]Suppose a function $ f:$ $ N \rightarrow N$, $ ff(n) \equal{} kn , k \in R$
Prove that
$ \frac {2k}{k \plus{} 1}n \leq f(n) \leq \frac {k \plus{} 1}{2}n$
And find an example of $ f(n)$ when $ k \equal{} 3$[/quote]

I think this is wrong.

For $ k\equal{}3$, for example, define $ f(n)$ as :

If $ n\equal{}3^p5^{2q}r$ (with $ \gcd(r,15)\equal{}1$), $ f(n)\equal{}5n\equal{}3^p5^{2q\plus{}1}r$

If $ n\equal{}3^p5^{2q\plus{}1}r$ (with $ \gcd(r,15)\equal{}1$), $ f(n)\equal{}\frac{3n}{5}\equal{}3^{p\plus{}1}5^{2q}r$

Clearly $ f(f(n))\equal{}3n$ and we never have $ \frac {3n}{2} \leq f(n) \leq 2n$


If x represents a two-digit positvie integer, compute the number of values of x such taht x^2+3x+2 is divisible by 6.

[quote="Phelpedo"][hide]Factor: $(x+1)(x+2)$. One of these is always even, so you just have to find all two digit numbers congruent to 1 or 2 mod 3.[/hide][/quote]

(x+1)/6 = n, where n is an integer
x+1=6n
x= 6n -1

Or,

(x+2)/6 = n, where n is an integer
x+2 = 6n
x=6n-2

I think you forgot the case where (x+1)|3 and (X+2)|2 or vice versa

A small error, neelnal. Note that the product of 2 numbers can be divisible by 6 without either of those two numbers being divisible by 6. The most basic example of this: 2*3 = 6.

[hide="I think its time we got the answer up here"]
all positive integers congruent to 1 or 2 mod 3 satisfy the eqn
so 66 of the first 99 positive integers work
6 of those are single digit
so our answer should be [b]60[/b][/hide]

Ooh. Ok yes. I'm sorry I made that mistake!

Stupid me!! :rotfl:

[hide="hint #1"]Don't forget to factor![/hide]

[hide="hint #2"]What in general are we looking for?[/hide]

[hide="hint #3"]Use modulo arithmatic.[/hide]

[hide="solution"]We can factor the equation to get (x+1)(x+2). We are assured that the product of those two is even. We need that product to also be a multiple of three. The only time that can't be reached is when x is a multiple of three. Otherwise, the first equation is a multiple of 6. There are 33 of the first 99 natural numbers that are multiples of three. 3 of those are single digit multiples. Since there are 90 two digit numbers, there are 90-30=[b]60[/b] values of x that make the first equation a multiple of 6.[/hide]


CBSE is probably trying to take the load off its database with 10 lakh people trying to find their results..

Hence realasing the results a week earlier.. :lol:

Haha. I am sure 90% of delhi doesnt know about this since they are gone to give CEE today. :p

c dis http://www.hindu.com/2009/05/30/stories/2009053058900300.htm

is it true results will cum at 2 pm

air 1124
wow man
i am pretty damn consistent :)

hey i was getting lesser marks than u..how can i get AIR 778??

State rank 88..nice :P

People in Delhi really perform well.

I got     AIR 1151
   State rank 41 (  :P Not much competition in chennai compared to Delhi)

Tamil Nadu state 1 is V Prashant( AIR 8  :w00t:  )

I think except chennai there is not much knowledge among students about engineering entrances. all they aim is state boards and get into local colleges.

air 74
state 3  (TN )
marks 377

how much did prashanth get

and whos air 1

Nitin jain got 417.

AIR 3(Shyam Upadhyay of Chandigarh..nice friend of mine :oops: ) got 409


Debra began playing a computer game at 6:52 p.m. and she finished
at 7:37 p.m. the same day. How many minutes over her 30-minute
limit did she play?

We can see that her limit would give her a stop at 7:22 pm. But 7:37 pm is $ \boxed{15}$ over this limit.

[b]another way to do this is by

beginning - 6:52 +30=6:82       82-60=22

so she should have stopped by 7 : 22 but she stopped at 7 : 3 7 so subtracting we get the answer  15 or $\boxed{15}$ [/b]


Prove that (a+b)(a+c) is bigger than or equals to 2 * the root of abc(a+b+c) when a, b and c are real, positive numbers.

I don't know how to type that in one line with square roots and other things, how do I do that?

And is there a place where I can find some inequalities to practice on? I'm not so good at them.

Regarding your inequality the case when $a=0$ is trivial so if $a\neq 0$ then the  inequality is equivalent to:
\[ 2\sqrt{\alpha\beta(1+\alpha+\beta)}\leq (1+\alpha)(1+\beta) \]
$\alpha,\beta\geq 0$ and in this case you just have to apply the lemma $2\sqrt{ab}\leq (a+b)/2$ by considering $a=\alpha\beta$ and $b=1+\alpha+\beta$ you'll have the result directly ;).
Now for ressources related to inequalities ,there are several ones on the web if you have ghostview then you can check the packet on inequalities on this link :) :
http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html

How do you know that they are equivalent? Sorry, I'm new when it comes to proving inequalities. And I can't open the inequality packets, do I need a program to open them?

[quote="MoHaMeD"]Regarding your inequality the case when $a=0$ is trivial so if $a\neq 0$ then the  inequality is equivalent to:
\[ 2\sqrt{\alpha\beta(1+\alpha+\beta)}\leq (1+\alpha)(1+\beta)  \]
$\alpha,\beta\geq 0$ and in this case you just have to apply the lemma $2\sqrt{ab}\leq (a+b)$ by considering $a=\alpha\beta$ and $b=1+\alpha+\beta$ you'll have the result directly.[/quote]
if $ a \neq 0$  then divide the two terms of the inequality by $a$ take $\alpha=b/a$ and $\beta=c/a$ this is what i meant by "equivalent".
for the packet yes you need $Ghostview$ to view it try to ask a teacher of yours who's interested in surfing the web i guess he should have it.

[quote="Masamune"]$(a+b)(a+c) \ge 2\sqrt{abc(a+b+c)} \ \ a, b, c \in \mathbb{R^+}$[/quote]$\frac{(a+b)(a+c)}{2} \ge \sqrt{abc(a+b+c)}$ expanding, $\frac{a^2+ab+ac+bc}{2} \ge \sqrt{abc(a+b+c)}$, factoring, $\frac{a(a+b+c)+bc}{2} \ge \sqrt{abc(a+b+c)}$ which is true by AM-GM.


Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that:

 $f(x+y)-f(x-y) = 4xy$ for all $x$ and $y$ non zero  in $\mathbb{R}$

[quote="lordWings"]Too easy or am I wrong?
[hide="Hidden text"]$x=y=\frac{t}{2}$ $\to$ $f(t)=t^2+f(0)$ for every $t$ non zero in $\mathbb{R}$.

So the only solutions are $f(t)=t^2+C$ for any $C$ in $\mathbb{R}$.[/hide][/quote]

Good Response if $f$ is defined in zero. But it is not!
 Try again  :oops:

[quote="afaouzi"]$f : \mathbb{R} \to \mathbb{R}$[/quote]
[quote="afaouzi"]Good Response if $f$ is defined in zero. But it is not![/quote]
So it's $f : \mathbb{R^*} \to \mathbb{R}$. OK, new problem. :P I'll try later.

$x=\frac{z+t}{2}$, $y=\frac{z-t}{2}$  $\to$ $f(z)-f(t)=z^2-t^2$ for every $t,z\in\mathbb{R^*}$.
$z$ and $t$ are independent, so we must have $f(z)=z^2+C$ for every $z\in\mathbb{R^*}$.


induction
combinatorics unsolved
combinatorics

as such the equality follows almost trivially by induction, but here's a combinatorial argument as well:(we prove
$(m+1)\sum_{j=1}^n\prod_{k=1}^m(j+k-1)=(n+m)(n+m-1)\cdots(n+1)n$)
 the RHS counts precisely the number of $m+1$-tuples from the set $\{1,2,\ldots,m+n\}$ with all entries pairwise distinct.as for the LHS, we count the number of tuples with $j+m$ as its largest entry in the tuple, where $j$ runs from $1$ to $n$. since there are $m+1$ positions where $j+m$ can occur and $ (\prod_{k=1}^m (j+k-1)$ such tuples, this is precisely the LHS.


function
trigonometry
real analysis
real analysis unsolved

Is the set of functions $ K \equal{} \{\cos nt: n\in N\}$ compact or closed, or both? (in $ C^0([0,1],\mathbf{R})$, under norm $ |f|\equal{}\sup_{t\in[0,1]} f(t)$)

Does the sequence $ \cos nt$ have a convergent subsequence?


calculus
integration
geometry
rectangle
complex analysis
complex analysis unsolved

$ \int_0^\infty \frac{e^{i \beta x}}{e^x \minus{} e^{\minus{}x}} dx \equal{} ?$

The integral may be written as

$ \frac{1}{2}\int_{0}^{\infty} e^{i\beta x}\sinh x dx$

which may be integrated by parts (twice).

No it can't because $ \sinh x \neq \frac{2}{e^x \minus{} e^{\minus{}x}}$.

But it is equal to $ \frac{e^{x} \minus{} e^{\minus{}x}}{2}$.

Yeah, $ \frac{e^{x}\minus{}e^{\minus{}x}}{2} \equal{} \sinh x$ but in integral there is: $ \int \frac{e^{i \beta x}}{2 \sinh x} dx$

Oops! Back to the drawing board.

I presume you've considered  the rectangle with vertices at $ \minus{}R$, $ R$, $ R\plus{} \pi i$ and $ \minus{}R \plus{} \pi i$ and the residue theorem.

[quote="@petko"]$ \int_0^\infty \frac {e^{i \beta x}}{e^x \minus{} e^{ \minus{} x}} dx \equal{} ?$[/quote]

I think the integrand is asymptotic to $ \frac {1}{x}$ near zero and so diverge.


Given n possitive real numbers $a_1;a_2;...;a_n$
Prove the ineq :
$(\sqrt{2})^n(a_1+a_2)(a_2+a_3)...(a_n+a_1)<(a_1+a_2+a_3)(a_2+a_3+a_4)...(a_n+a_1+a_2)$
Can someone replace $\sqrt{2}$ by another number for the best ineq ?
I think it is 1.5 but I can't prove it :(

Just note that $(a_1+a_2+a_3)^2\geq(2a_1+a_2)(2a_3+a_2)$ and $(2a_1+a_2)(2a_2+a_1)\geq 2(a_1+a_2)^2$

By the way, I believe I just saw this problem 2~3 weeks ago in a competition in Taiwan. Did you get this problem from a friend in Taiwan?

P.S. This solution is not mine. One of my friends gave me it that time.

Dear Darkseer! I get this problem in a old Russian MO.
$(2a_1+a_2)(a_1+2a_2)=2a_1^2+2a_2^2+4a_1a_2+a_1a_2>2(a_1+a_2)^2$
The equal didn't exist in that ineq. So we can only prove the base problem but I want to get a stronger ineq. For $a_1=a_2=...=a_n$ we'll get $A\le (1,5)^n$

I conjecture that the best $A$ for odd $n$ is $A=\frac{3^n}{2^n}$ for $n=3$ and $n=5$, and $A=2^{(n+1)/2}$ for odd $n\ge7$.
Can anybody to prove or disprove this one for $n=5$?

[quote="Vasc"]A nice inequality is also the following:

$(1+\frac{2a}{b+c})(1+\frac{2b}{c+d})(1+\frac{2c}{d+a})(1+\frac{2d}{a+b}) \ge 9.$

I wait a nice solution. :)[/quote]
Do we have any conditions on $a, b, c, d$?

In the given context, $a,b,c,d$ are clearly non-negative.

Very nice problem.
I'm very glad to find out a short solution only use AM-GM!

Post your solution, Hungkhtn.
And think to the stronger inequality
$(1+\frac{3a}{b+c})(1+\frac{3b}{c+d})(1+\frac{3c}{d+a})(1+\frac{3d}{a+b}) \ge 16.$

$\displaystyle (1+\frac{2a}{b+c})(1+\frac{2b}{c+d})(1+\frac{2c}{d+a})(1+\frac{2d}{a+b}) \ge 9$

clearly,it's equal with:

$\displaystyle (1+\frac{a+c}{a+b})(1+\frac{a+c}{c+d})(1+\frac{b+d}{b+c})(1+\frac{b+d}{a+d}) \ge 9$

And easy to prove:

$\displaystyle (1+\frac{a+c}{a+b})(1+\frac{a+c}{c+d}) \ge (1+\frac{2a+2c}{a+b+c+d})^2$

$\displaystyle (1+\frac{b+d}{b+c})(1+\frac{b+d}{a+d}) \ge (1+\frac{2b+2d}{a+b+c+d})^2$

By AM-GM
Let
$\displaystyle x=\frac{a+c}{a+b+c+d}$
$\displaystyle y=\frac{b+d}{a+b+c+d}$

Then $x+y=1$ and $0 \le x,y \le 1$

$\displaystyle (1+2x)(1+2y) \ge 3 \leftrightarrow 1+2(x+y)+xy \ge 3$

True,and the equal hold with  $a=c=0,b=d$ or $b=d=0,a=c$

Vasc,can your post your solution?

What do you mean "later" for you, Vasc ?

A day, a week, a month, a year, .........

Thank you. :)

A month, Manlio.
But I hope somebody give a solution in the meantime.

[quote="Vasc"]Yes, but later, because my solution is also valid for the inequality

$(1+\frac{3a}{b+c})(1+\frac{3b}{c+d})(1+\frac{3c}{d+a})(1+\frac{3d}{a+b}) \ge 16.$

Very nice your solution for the previous inequality![/quote]

Can we replace 3 to other number,4 or 5?
Is your solution enough to complete this problem,Vasc?

Yes, we can replace 3 with 4 or 5.
My solution works for all these cases.

Ok,I've just found the general ineq:

$(1+\frac{ka}{b+c})(1+\frac{kb}{c+d})(1+\frac{kc}{d+a})(1+\frac{kd}{a+b}) \ge (1+k)^2$

VASE :can you show the solution to
$\sum\frac{a}{b+c} \ge 1$.

and

$\sum\frac{ab}{(b+c)(c+d)} \ge 1$.

I have some difficult in solving these.thank you in advance

[quote="Vasc"]Of course, for $k$ positive.
It follows from well known inequality
$\sum\frac{a}{b+c} \ge 1$.

and from

$\sum\frac{ab}{(b+c)(c+d)} \ge 1$.

Congratulations![/quote]

Ok,My solution is that.But it should be:
$\sum\frac{a}{b+c} \ge 2$.

easy by Cauchy-Swaczer.

And the rest inequality:

$\sum\frac{ab}{(b+c)(c+d)} \ge 1$.

I proof it by expanding.I can't sovle it by AM-Gm or Cauchy!

Vasc,I want to see your solution for it,if it doesn't  use expanding?

I didn't use expanding.
I used the inequalities
$x_1+x_3 \ge 2x_1x_3$,
$x_2+x_4 \ge 2x_2x_4$,
where $x_1=a/(b+c)$ etc.

Can you please explain better your idea, Vasc? :?

For the first ineq you have

$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \geq 2\frac{ac}{(b+c)(d+a)}+2\frac{bd}{(c+d)(a+b)}$

which is not greater than 2 ?????????????????????????????

And for the second I really don't see how to apply your idea.!!!!!!!!!!!!!!!!!!

Inequality
\[ \left( \frac{a_1 + a_2 + a_3}{a_1 + a_2} \right)
   \left( \frac{a_2 + a_3 + a_4}{a_2 + a_3} \right) \dots
   \left( \frac{a_n + a_1 + a_2}{a_n + a_1} \right) \geq
   \left( \! \sqrt{2} \, \right)^n \]
is not from old Russian MO, it's my problems from 239 school MO, 2000. For
an even n the constant $\sqrt{2}$ in the best. For odd it can be improved.
We can put constant 3/2 for n=3 and n=5 only. Case n=5 is my problems from
239 school MO, 2000. The best constnt for odd numbers greater than 5 is
unknown. Details you can see at my articles "About Mongolian Inequality",
they are published at journal "Matematichaskoe Prosveschenie", 2003
(condensed version) and at Problems of SPb MO, 2002 (full version). You also
can find articles on my www. Unfortunately they are in Russian only.

[quote="Vasc"]I think the following conjecture is valid for $n=5$ and $p \le 5/4$:

$\prod_{cyclic}(1+\frac{pa_1}{a_2+a_3}) \ge (1+p/2)^5$.[/quote]

Very nice!And you have found a nice solution,Vasc?

[quote="hungkhtn"]
...
And easy to prove:

$\displaystyle (1+\frac{a+c}{a+b})(1+\frac{a+c}{c+d}) \ge (1+\frac{2a+2c}{a+b+c+d})^2$

$\displaystyle (1+\frac{b+d}{b+c})(1+\frac{b+d}{a+d}) \ge (1+\frac{2b+2d}{a+b+c+d})^2$
...
[/quote]

Can You show proof?

[quote="Vasc"]I think the following conjecture is valid for $ n \equal{} 5$ and $ p \le 5/4$:

$ \prod_{cyclic}(1 \plus{} \frac {pa_1}{a_2 \plus{} a_3}) \ge (1 \plus{} p/2)^5$.[/quote]

Have you proved it？