## IMO 1995-2

Discussion on International Mathematical Olympiad (IMO)
Hasib
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### IMO 1995-2

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:
A man is not finished when he's defeated, he's finished when he quits.

Masum
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Joined: Tue Dec 07, 2010 1:12 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 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 $\sum \frac {x^2} {y+z}\ge \sum \frac {x^2} {x+x}=\sum \frac x 2\ge 3\frac {\sqrt {xyz}} 2=\frac 3 2$
One one thing is neutral in the universe, that is $0$.