Specific inequality Let $x,y,z$ be different real numbers . Prove that:
$$\frac{x^2y^2+1}{(x-y)^2}+\frac{y^2z^2+1}{(y-z)^2}+\frac{z^2x^2+1}{(x-z)^2} \geq \frac{3}{2}$$
 A: 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
A: By AM-GM
$$\sum_{cyc}\frac{x^2y^2+1}{(x-y)^2}=\sum_{cyc}\frac{x^2y^2}{(x-y)^2}+\sum_{cyc}\frac{1}{(x-y)^2}\geq2\sqrt{\sum_{cyc}\frac{x^2y^2}{(x-y)^2}\sum_{cyc}\frac{1}{(x-y)^2}}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{x^2y^2}{(x-y)^2}\sum_{cyc}\frac{1}{(x-y)^2}\geq\frac{9}{16}$$
or
$$\left(\sum_{cyc}(x^2y+x^2z-2xyz)\right)^2\sum_{sym}\left(7x^4y^2-5x^3y^3-5x^4yz+2x^3y^2z+x^2y^2z^2\right)\geq0.$$
Hence, it remains to prove that
$$\sum_{sym}\left(7x^4y^2-5x^3y^3-5x^4yz+2x^3y^2z+x^2y^2z^2\right)\geq0.$$
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Hence, $$\sum\limits_{sym}x^4y^2=(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)-3x^2y^2z^2=$$
$$=-3w^6+A_1(u,v^2)w^3+B_1(u,v^2);$$
$$\sum_{cyc}x^3y^3=27v^6-27uv^2w^3+3w^6;$$
$$\sum_{cyc}x^4yz=27u^3w^3-27uv^2w^3+3w^6$$ and
$$\sum_{cyc}x^3y^2z=9uv^2w^3-3w^6.$$
Thus, our inequality it's just $f(w^3)\geq0$, where $f$ is a concave function,
which says that it's enough to prove our inequality for an extremal value of $w^3$,
which happens for equality case of two variables.
Since the last inequality is homogeneous and even degree, 
it's enough to assume $y=z=1$, which gives $(x-1)^4\geq0$.
Done!
