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

Question

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

Answer 1

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 $(a+b+c-1)^2 \geqslant 0$

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