Prove that $\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1$ Question -

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

My work -
first i apply directly C-S and after simplification i have to prove that 
$4(xy+yz+zx)>x^2+y^2+z^2$ which i am not able to prove ..
then multiplying $x^2,y^2,z^2$ to numerators and denominators of corresponding fractions respectively i again apply C-S and this time we have to prove that after simplification
$(xy)^2+(yz)^2+(zx)^2 + 2x^3y+2y^3z+2z^3x > 0$ which again i fail to prove ...
I try to do with other inequalities but none of them working.
any help will be helpful
thankyou

Let $a = \dfrac{x}{x-y}, b = \dfrac{y}{y-z}, c = \dfrac{z}{z-x}$, and note 
$a+b+c = ab+bc+ca+1$.  We need to show $a^2+b^2+c^2 \geqslant 1$.  But this is now just equivalent to $(a+b+c-1)^2 \geqslant 0$ 

Alternately, the original inequality $\iff \dfrac{(x^2y+y^2z+z^2x-3xyz)^2}{(x-y)^2(y-z)^2(z-x)^2}\geqslant 0$
