<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] I think our inequality is true for all reals $x$, $y$ and $z$ such that $(x-y)(x-z)(y-z)\neq0$. 
Indeed, if $z=0$ then
$$\sum_{cyc}\frac{(2x-y)^2}{(x-y)^2}=\frac{(2x-y)^2}{(x-y)^2}+5\geq5.$$
Let $xyz\neq0$, $\frac{2x-y}{x-y}=a$, $\frac{2y-z}{y-z}=b$ and $\frac{2z-x}{z-x}=c$.
Thus, $(2-a)x=(1-a)y$, $(2-b)y=(1-b)z$ and $(2-c)z=(1-c)x$, which gives
$$(2-a)(2-b)(2-c)xyz=(1-a)(1-b)(1-c)xyz$$ or
$$(2-a)(2-b)(2-c)=(1-a)(1-b)(1-c)$$ (now, try to prove the starting inequality by yourself without to see the rest) or
$$7-3(a+b+c)+ab+ac+bc=0$$ or
$$14-6(a+b+c)+2(ab+ac+bc)=0$$ or
$$5+9-6(a+b+c)+(a+b+c)^2=a^2+b^2+c^2$$ or
$$a^2+b^2+c^2=5+(a+b+c-3)^2,$$
which gives $$a^2+b^2+c^2\geq5$$ and we are done!
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