# Specific inequality

Let $x,y,z$ be different real numbers . Prove that: $$\frac{x^2y^2+1}{(x-y)^2}+\frac{y^2z^2+1}{(y-z)^2}+\frac{z^2x^2+1}{(x-z)^2} \geq \frac{3}{2}$$

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Maybe you think a=x,b=y,c=z –  Adi Dani Dec 16 '12 at 16:12
sorry i 've check –  Haruboy15 Dec 16 '12 at 16:15
I find the min happen when $z=0$ and $x=-y$ –  Haruboy15 Dec 16 '12 at 23:51

Hint : We have Dao Hai Long inequality : $$\sum \frac{a+b}{a-b}.\frac{b+c}{b-c}=-1$$ So : $$\sum \frac{(a+b)^2}{(a-b)^2} \geq 2$$ So: $$\sum \frac{ab}{(a-b)^2} \geq \frac{-1}{4} (1)$$ Then : $$\sum \frac{1-ab}{a-b}=-1$$ So: $$\sum \frac{(1-ab)^2}{(a-b)^2} \geq 2 (2)$$ From (1) and (2) we have solution