# Let (x,y,z) in R, prove that ((2x-y)/(x-y))^2+((2y-z)/(y-z))^2 +((2z-x)/(z-x))^2 >= 5

Let $\left(x,y,z\right)\in \mathbb{R}$, prove that $\left(\frac{2x-y}{x-y}{\right)}^{2}+\left(\frac{2y-z}{y-z}{\right)}^{2}+\left(\frac{2z-x}{z-x}{\right)}^{2}⩾5$
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smh3402en
$\begin{array}{r}\left(2{y}^{2}z+2{z}^{2}x+2{x}^{2}y-y{z}^{2}-z{x}^{2}-x{y}^{2}-3xyz{\right)}^{2}\ge 0\end{array}$
This can be expanded out to give
$\begin{array}{r}4\sum _{cyc}{y}^{2}{z}^{4}-4\sum _{cyc}{y}^{3}{z}^{3}+\sum _{cyc}{y}^{4}{z}^{2}-4\sum _{cyc}xy{z}^{4}+\\ 14\sum _{cyc}x{y}^{2}{z}^{3}-10\sum _{cyc}x{y}^{3}{z}^{2}-3{x}^{2}{y}^{2}{z}^{2}\ge 0\end{array}$
This can be rearranged to
$\begin{array}{r}\left(2x-y{\right)}^{2}\left(y-z{\right)}^{2}\left(z-x{\right)}^{2}+\left(x-y{\right)}^{2}\left(2y-z{\right)}^{2}\left(z-x{\right)}^{2}+\left(x-y{\right)}^{2}\left(y-z{\right)}^{2}\left(2z-x{\right)}^{2}\ge 5\left(x-y{\right)}^{2}\left(y-z{\right)}^{2}\left(z-x{\right)}^{2}\end{array}$
Now divide by
$\left(x-y{\right)}^{2}\left(y-z{\right)}^{2}\left(z-x{\right)}^{2}$ and we have
$\begin{array}{r}\frac{\left(2x-y{\right)}^{2}}{\left(x-y{\right)}^{2}}+\frac{\left(2y-z{\right)}^{2}}{\left(y-z{\right)}^{2}}+\frac{\left(2z-x{\right)}^{2}}{\left(z-x{\right)}^{2}}\ge 5.\end{array}$
###### Did you like this example?
flatantsmu
This is a special case of the following inequality:
$\sum _{a,b,c}\left(\frac{ma-nb}{a-b}{\right)}^{2}\ge {m}^{2}+{n}^{2},$
where $m,n\in \mathbb{R}$ and $a,b,c$ are distinct reals.
The proof is simply completing the square: start by writing
$\frac{ma-nb}{a-b}=m+\left(m-n\right)\frac{b}{a-b}$
and
$\frac{mc-na}{c-a}=n+\left(m-n\right)\frac{c}{c-a}$
and leave the middle term untouched first.
Let me know if still need further assistance.
EDIT: Completing the hint:
$\sum _{a,b,c}\left(\frac{ma-nb}{a-b}{\right)}^{2}-{m}^{2}-{n}^{2}=\left(\frac{mb-nc}{b-c}{\right)}^{2}+\left(m-n{\right)}^{2}\left(\frac{{b}^{2}}{\left(a-b{\right)}^{2}}+\frac{{c}^{2}}{\left(c-a{\right)}^{2}}\right)+2\left(m-n\right)\left(\frac{mb}{a-b}+\frac{nc}{c-a}\right)=$
$=\left(\frac{mb-nc}{b-c}{\right)}^{2}+\left(m-n{\right)}^{2}\left(\frac{{b}^{2}}{\left(a-b{\right)}^{2}}+\frac{{c}^{2}}{\left(c-a{\right)}^{2}}\right)+2\left(m-n\right)\frac{bc\left(m-n\right)-a\left(mb-nc\right)}{\left(a-b\right)\left(c-a\right)}=$
$=\left(\frac{mb-nc}{b-c}{\right)}^{2}+\left(m-n{\right)}^{2}\left(\frac{{b}^{2}}{\left(a-b{\right)}^{2}}+\frac{{c}^{2}}{\left(c-a{\right)}^{2}}+\frac{2bc}{\left(a-b\right)\left(c-a\right)}\right)-\frac{2a\left(mb-nc\right)\left(m-n\right)}{\left(a-b\right)\left(c-a\right)}=$
$=\left(\frac{mb-nc}{b-c}{\right)}^{2}+\left(m-n{\right)}^{2}\left(\frac{b}{a-b}+\frac{c}{c-a}{\right)}^{2}-\frac{2a\left(m-n\right)\left(mb-nc\right)}{\left(a-b\right)\left(c-a\right)}=$
$\left(\frac{mb-nc}{b-c}{\right)}^{2}+\left(m-n{\right)}^{2}\left(\frac{a\left(c-b\right)}{\left(a-b\right)\left(c-a\right)}{\right)}^{2}-\frac{2a\left(m-n\right)\left(mb-nc\right)}{\left(a-b\right)\left(c-a\right)}=$
$=\left(\frac{mb-nc}{b-c}-\frac{a\left(m-n\right)\left(c-b\right)}{\left(a-b\right)\left(c-a\right)}{\right)}^{2}\ge 0.$