Let $x,y,z\in \left[ 0,\infty \right) $. Show that, for all $n\in \mathbb{N}$, we have $$ \sum_{cyc}x^{n}\left( y+z-2x\right) \leq 0. $$
First of all, I made the multiplication to obtain $$ \sum_{cyc}\left( x^{n}y+x^{n}z-2x^{n+1}\right) \leq 0\Leftrightarrow \sum_{cyc}x^{n}y+x^{n}z\leq 2\left( x^{n+1}+y^{n+1}+z^{n+1}\right) . $$ Hence $$ x\left( y^{n}+z^{n}\right) +y\left( z^{n}+x^{n}\right) +z\left( x^{n}+y^{n}\right) \leq 2\left( x^{n+1}+y^{n+1}+z^{n+1}\right) . $$ My hope was to prove that $$ x\dfrac{y^{n}+z^{n}}{2}\leq x^{n+1}\Leftrightarrow \dfrac{y^{n}+z^{n}}{2} \leq x^{n},x\not=0. $$ But this is not true, since for $x=1,y=2$ and $z=3$ is false. Am I wrong with something? Or how can I prove this inequality?
It's $$\sum_{cyc}(2x^{n+1}-x^ny-x^nz)\geq0$$ or $$\sum_{cyc}(x^{n+1}-x^ny-xy^n+y^{n+1})\geq0$$ or $$\sum_{cyc}(x^n-y^n)(x-y)\geq0,$$ which is obvious.