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

$$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2-2=\frac{\left(\sum\limits_{cyc}(x^3-x^2y-x^2z+xyz)\right)^2}{\prod\limits_{cyc}(x-y)^2}\geq0.$$
Also, we can use the following way.
Since $$\sum_{cyc}\left(\frac{x}{y-z}\cdot\frac{y}{z-x}\right)=\sum_{cyc}\frac{xy(x-y)}{\prod\limits_{cyc}(x-y)}=\frac{(x-y)(x-z)(y-z)}{(x-y)(y-z)(z-x)}=-1,$$ we obtain
$$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2-2=\left(\sum_{cyc}\frac{x}{y-z}\right)^2\geq0.$$
The last solution it's just your work:
$$p^2+q^2+r^2=p^2+q^2+r^2+2(pq+pr+qr)+2=(p+q+r)^2+2\geq2.$$
The equality occurs for example, for $x=-y$ and $z=0$. 
