Proving expression$\geq 2$, where variables are $x,y,z$

Question

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

Answer 1

$$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2-2=\frac{\left(\sum\limits_{cyc}(x^3-x^2y-x^2z+xyz)\right)^2}{\prod\limits_{cyc}(x-y)^2}\geq0.$$ 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 $z=0$.

Answer 2

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 $$\sum \frac{x^2}{(y-z)^2} \geqslant \frac{\left[x(2x-y-z)+y(2y-z-x)+z(2z-x-y)\right] ^2}{(y-z)^2(2x-y-z)^2+(z-x)^2(2y-z-x)^2+(x-y)^2(2z-x-y)^2}=2.$$