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)

This is a special case of the following inequality: $$\sum_{a,b,c}\Big(\dfrac{ma-nb}{a-b}\Big)^2\geq m^2+n^2,$$
where $m,n\in\mathbb{R}$ and $a,b,c$ are distinct reals. 
The proof is simply completing the square: start by writing $$\dfrac{ma-nb}{a-b} = m+(m-n)\dfrac{b}{a-b}$$ and $$\dfrac{mc-na}{c-a} = n+(m-n)\dfrac{c}{c-a}$$ and leave the middle term untouched first. 
Let me know if still need further assistance. 
EDIT: Completing the hint: $$\sum_{a,b,c}\Big(\dfrac{ma-nb}{a-b}\Big)^2-m^2-n^2 = \Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}\Big)+2(m-n)\Big(\dfrac{mb}{a-b}+\dfrac{nc}{c-a}\Big)=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}\Big)+2(m-n)\dfrac{bc(m-n)-a(mb-nc)}{(a-b)(c-a)}=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}+\dfrac{2bc}{(a-b)(c-a)}\Big)-\dfrac{2a(mb-nc)(m-n)}{(a-b)(c-a)}=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b}{a-b}+\dfrac{c}{c-a}\Big)^2 - \dfrac{2a(m-n)(mb-nc)}{(a-b)(c-a)}=$$
$$\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{a(c-b)}{(a-b)(c-a)}\Big)^2 - \dfrac{2a(m-n)(mb-nc)}{(a-b)(c-a)}=$$ $$=\Big(\dfrac{mb-nc}{b-c} - \dfrac{a(m-n)(c-b)}{(a-b)(c-a)}\Big)^2\geq 0.$$
