Proving expression$\geq 2$, where variables are $x,y,z$ 
If $x,y,z$ are distinct real number. Then prove that $$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2\geq 2$$

Try : let $\displaystyle \frac{x}{y-z}=p\Rightarrow x=py-pz\Rightarrow x-py+pz=0$
And let  $\displaystyle \frac{y}{z-x}=q\Rightarrow y=qz-qx\Rightarrow qx+y-qz=0$
let $\displaystyle \frac{z}{x-y}=r\Rightarrow z=xr-yr\Rightarrow xr-yr-z=0$
So $$\begin{bmatrix}
 1& -p &p \\ 
 q& 1 & -q\\ 
 r& -r & -1
\end{bmatrix}\begin{bmatrix}
x\\ 
y\\ 
z
\end{bmatrix}=\begin{bmatrix}
0\\ 
0\\ 
0
\end{bmatrix}$$
Which is $AX=B$
So for solution of equation , $|A|=pq+qr+rp+1=0$
Now using $(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$
So $$p^2+q^2+r^2\geq 2$$
What i have used above is right or wrong. If it is right then i did not understand when will the equality hold
If it is wrong then could someone explain me a right solution, thanks

Note
$$ \sum x(2x-y-z)=2(x^2+y^2+z^2-xy-yz-zx),$$
and
$$\sum (y-z)^2(2x-y-z)^2=2(x^2+y^2+z^2-xy-yz-zx)^2.$$
Therefore, according to the Cauchy-Schwarz inequality we have
$$\sum \frac{x^2}{(y-z)^2} \geqslant \frac{\left[x(2x-y-z)+y(2y-z-x)+z(2z-x-y)\right] ^2}{(y-z)^2(2x-y-z)^2+(z-x)^2(2y-z-x)^2+(x-y)^2(2z-x-y)^2}=2.$$
