Minimum value of rational expression 
If $x,y,z$ are distinct real number, Then minimum value of   $\displaystyle \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2$

Try: $$\left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2\geq \frac{x^2+y^2+z^2}{(x-y)^2+(y-z)^2+(z-x)^2}$$
could some help me to solve this, thanks
 A: If $z=0$ and $x=-y$ then we get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{x^2}{(y-z)^2}\geq2.$$
Let $y=x+u$ and $w=x+v$.
Hence, we need to prove that
$$\frac{x^2}{(u-v)^2}+\frac{(x+u)^2}{v^2}+\frac{(x+v)^2}{u^2}\geq2$$ or
$$\left(\frac{1}{(u-v)^2}+\frac{1}{u^2}+\frac{1}{v^2}\right)x^2+2\left(\frac{u}{v^2}+\frac{v}{u^2}\right)x+\frac{u^2}{v^2}+\frac{v^2}{u^2}-2\geq0,$$ for which it's enough to prove that
$$\left(\frac{u}{v^2}+\frac{v}{u^2}\right)^2-\left(\frac{1}{(u-v)^2}+\frac{1}{u^2}+\frac{1}{v^2}\right)\left(\frac{u^2}{v^2}+\frac{v^2}{u^2}-2\right)\leq0$$ or
$$\frac{(u+v)^2(u^2-uv+v^2)^2}{u^4v^4}\leq\frac{(u^2v^2+(u-v)^2(u^2+v^2))(u^2-v^2)^2}{(u-v)^2u^4v^4}$$ or
$$(u^2-uv+v^2)^2\leq u^2v^2+(u-v)^2(u^2+v^2)$$ or
$$(u^2-uv+v^2)^2-u^2v^2\leq(u-v)^2(u^2+v^2)$$ or
$$(u-v)^2(u^2+v^2)\leq(u-v)^2(u^2+v^2).$$
Done!
A: we have to prove that $$ \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2\geq 2$$
this is true since this inequality is equivalent to $${\frac { \left( {x}^{3}-y{x}^{2}-z{x}^{2}-{y}^{2}x+3\,zyx-{z}^{2}x+{y}
^{3}-z{y}^{2}-{z}^{2}y+{z}^{3} \right) ^{2}}{ \left( y-z \right) ^{2}
 \left( -z+x \right) ^{2} \left( x-y \right) ^{2}}}
\geq 0$$
