Prove that $$x^t(x-y)(x-z) + y^t(y-z)(y-x) + z^t(z-x)(z-y) \geq 0$$, if $t \in \mathbb{R} $ and $x,y,z$ are positive numbers.
This inequality was given to me by my friend. I tried various approaches like C-S and Holder, but failed. For $t>0$, the inequality is obvious by Schur but for case $t<0$, I am not able to find a suitable approach.
Any help will be appreciated.
Thanks in advance ! :)
We'll prove the stronger inequality.
Let $f$ is a positive convex function on $(0,+\infty)$. Prove that: $$\sum_{cyc}f(x)(x-y)(x-z)\geq0$$ Indeed, let $x\geq y\geq z$.
Hence, $f(x)\geq f(y)$ or $f(z)\geq f(y)$.
Let $f(x)\geq f(y)$.
Thus, $$\sum_{cyc}f(x)(x-y)(x-z)\geq f(x)(x-y)(x-z)+f(y)(y-x)(y-z)=$$ $$=(x-y)(f(x)(x-z)-f(y)(y-z))\geq(x-y)(f(x)(y-z)-f(y)(y-z))=$$ $$=(x-y)(y-z)(f(x)-f(y))\geq0$$ The second case is the same.
Thus, we proved the starting inequality for $t\geq1$ and for $t<0$.
Also we have the following.
If $f$ an increasing positive function so for all positives $x$, $y$ and $z$ $$\sum_{cyc}f(x)(x-y)(x-z)\geq0.$$ Proof.
Let $x\geq y\geq z$.
Hence, $$\sum_{cyc}f(x)(x-y)(x-z)\geq f(x)(x-y)(x-z)+f(y)(y-x)(y-z)\geq$$
$$\geq(x-y)(y-z)(f(x)-f(y))\geq0.$$
It gives a proof of starting inequality for $0 Done!