<s>[INST] Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$ 
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \geqslant 5.$$

This is an assignment I got from a teacher. I've been baffled by the negative sign in the denominator.
Could you please provide insight as to how would one resolve this inequality please? (Do not solve yet, I've yet to turn in this assignment, I just want to get some tips)
 [/INST] We want to minimize
$$
\begin{align}
&\underbrace{\left(\frac{mx-ny}{x-y}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{x+y}{x-y}\right)^2}+\underbrace{\left(\frac{my-nz}{y-z}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{y+z}{y-z}\right)^2}+\underbrace{\left(\frac{mz-nx}{z-x}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{z+x}{z-x}\right)^2}\tag{1}
\end{align}
$$
Letting
$$
u=\frac{x+y}{x-y}\quad v=\frac{y+z}{y-z}\quad w=\frac{z+x}{z-x}\tag{2}
$$
transforms minimizing $(1)$ into minimizing
$$
\left(\frac{m+n}2+\frac{m-n}2u\right)^2+\left(\frac{m+n}2+\frac{m-n}2v\right)^2+\left(\frac{m+n}2+\frac{m-n}2w\right)^2\tag{3}
$$
where
$$
\underbrace{\log\left(\frac{u-1}{u+1}\right)}_{\log\left(\frac yx\right)}+\underbrace{\log\left(\frac{v-1}{v+1}\right)}_{\log\left(\frac zy\right)}+\underbrace{\log\left(\frac{w-1}{w+1}\right)}_{\log\left(\frac xz\right)}=0\tag{4}
$$
Thus, for all variations that maintain $(4)$; i.e.
$$
\frac{\delta u}{u^2-1}+\frac{\delta v}{v^2-1}+\frac{\delta w}{w^2-1}=0\tag{5}
$$
we want to minimize $(3)$; i.e.
$$
\left(\frac{m+n}2+\frac{m-n}2u\right)\delta u+\left(\frac{m+n}2+\frac{m-n}2v\right)\delta v+\left(\frac{m+n}2+\frac{m-n}2w\right)\delta w=0\tag{6}
$$
Orthogonality requires that to satisfy $(6)$ for all variations that satisfy $(5)$ there be a $\lambda$ so that
$$
\begin{align}
\lambda
&=\left(u^2-1\right)\left(\frac{m+n}2+\frac{m-n}2u\right)\\
&=\left(v^2-1\right)\left(\frac{m+n}2+\frac{m-n}2v\right)\\
&=\left(w^2-1\right)\left(\frac{m+n}2+\frac{m-n}2w\right)
\end{align}\tag{7}
$$
That is, $u,v,w$ are roots of
$$
x^3+\frac{m+n}{m-n}x^2-x-\frac{m+n+2\lambda}{m-n}=0\tag{8}
$$
The sum of the roots of $(8)$ is
$$
-\frac{m+n}{m-n}\tag{9}
$$
and the sum of the squares of the roots of $(8)$ is
$$
\left(\frac{m+n}{m-n}\right)^2+2\tag{10}
$$
Applying $(9)$ and $(10)$ to $(3)$ yields that the minimum is
$$
3\frac{(m+n)^2}4-\frac{m^2-n^2}2\frac{m+n}{m-n}+\frac{(m-n)^2}4\left[\left(\frac{m+n}{m-n}\right)^2+2\right]=\bbox[5px,border:2px solid #C0A000]{m^2+n^2}\tag{11}
$$
</s>