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)
 A: 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: $$\sum_{a,b,c}\Big(\dfrac{ma-nb}{a-b}\Big)^2-m^2-n^2 = \Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}\Big)+2(m-n)\Big(\dfrac{mb}{a-b}+\dfrac{nc}{c-a}\Big)=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}\Big)+2(m-n)\dfrac{bc(m-n)-a(mb-nc)}{(a-b)(c-a)}=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}+\dfrac{2bc}{(a-b)(c-a)}\Big)-\dfrac{2a(mb-nc)(m-n)}{(a-b)(c-a)}=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b}{a-b}+\dfrac{c}{c-a}\Big)^2 - \dfrac{2a(m-n)(mb-nc)}{(a-b)(c-a)}=$$
$$\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{a(c-b)}{(a-b)(c-a)}\Big)^2 - \dfrac{2a(m-n)(mb-nc)}{(a-b)(c-a)}=$$ $$=\Big(\dfrac{mb-nc}{b-c} - \dfrac{a(m-n)(c-b)}{(a-b)(c-a)}\Big)^2\geq 0.$$
A: Start with
\begin{eqnarray*}
(2y^2z+2z^2x+2x^2y-yz^2-zx^2-xy^2-3xyz)^2 \geq 0
\end{eqnarray*}
This can be expanded out to give
\begin{eqnarray*}
4 \sum_{cyc} y^2 z^4 -4 \sum_{cyc} y^3 z^3 +\sum_{cyc} y^4 z^2 -4 \sum_{cyc} x y z^4 +\\14 \sum_{cyc} xy^2 z^3 -10 \sum_{cyc} x y^3 z^2 -3x^2 y^2 z^2 \geq 0
\end{eqnarray*}
This can be rearranged to 
\begin{eqnarray*}
(2x-y)^2(y-z)^2(z-x)^2+(x-y)^2(2y-z)^2(z-x)^2+(x-y)^2(y-z)^2(2z-x)^2 \geq 5(x-y)^2(y-z)^2(z-x)^2
\end{eqnarray*}
Now divide by $(x-y)^2(y-z)^2(z-x)^2$ and we have 
\begin{eqnarray*}
\frac{(2x-y)^2}{(x-y)^2}+\frac{(2y-z)^2}{(y-z)^2}+\frac{(2z-x)^2}{(z-x)^2} \geq 5.
\end{eqnarray*}
Edit: 
I worked my way backwards from the result, using the "reduce" algebra package. I then guessed a square that would "consume" some of the terms & got lucky !

(2*x-y)^2*(y-z)^2*(z-x)^2+(x-y)^2*(2*y-z)^2*(z-x)^2+(x-y)^2*(y-z)^2*(2*z-x)^2-5*(x-y)^2*(y-z)^2*(z-x)^2;
(2*(y^2*z+z^2*x+x^2*y)-(yz^2+zx^2+x*y^2)-3*xyz)^2;

A: 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
$$
3\frac{(m+n)^2}4-\frac{m^2-n^2}2\frac{m+n}{m-n}+\frac{(m-n)^2}4\left[\left(\frac{m+n}{m-n}\right)^2+2\right]=\bbox[5px,border:2px solid #C0A000]{m^2+n^2}\tag{11}
$$
A: 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!
