IMO 1995-2

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Hasib
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IMO 1995-2

Unread post by Hasib » Sun Jan 02, 2011 6:34 pm

let, $a,b,c$ are positive real number such that $abc=1$ .
Prove that \[\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \ge \frac{3}{2}\]


so easy, do urself and be more confident :)

Hints:
Do u know about $AM-GM$ inequality?
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Masum
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Re: IMO 1995-2

Unread post by Masum » Tue Feb 08, 2011 1:54 pm

Hasib wrote:let, $a,b,c$ are positive real number such that $abc=1$ . Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \ge \frac{3}{2}$


so easy, do urself and be more confident :)

Hints:
Do u know about $AM-GM$ inequality?
Well,here is a shorter solution
Let $a=\frac 1 x,b=\frac 1 y,c=\frac 1 z,xyz=1,$wlog $x\ge y\ge z$ and the sequences $x^2$ and $\frac 1 {y+z}$ are similarly sorted.So from rearrangement inequality \[\sum \frac {x^2} {y+z}\ge \sum \frac {x^2} {x+x}=\sum \frac x 2\ge 3\frac {\sqrt[3] {xyz}} 2=\frac 3 2\]
One one thing is neutral in the universe, that is $0$.

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