Prove $x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)>0$ Let $x,y,z>0$ and $n\in\mathbb{N}$, prove that:
$$x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)>0$$
Some idea? I have tried to develop the multiplication and certain assumptions buy I can't see it clearly. Note that this is not true if $x=y=z$ or $x=y=z=0$.
 A: It's wrong. Try $n=3$, $x=-1$ and $y=z=1$.
For positives variables it's true.
Indeed, let $x\geq y\geq z$.
Thus, $$\sum_{cyc}x^n(x-y)(x-y)\geq$$
$$\geq x^n(x-y)(x-z)+y^n(y-x)(y-z)=(x-y)(x^n(x-z)-y^n(y-z))\geq0$$
because $x^n\geq y^n$ and $x-z\geq y-z.$
By the same way we can prove the following inequality.

Let $x$, $y$ and $z$ be positive numbers and $t$ be a real number. Prove that:
  $$x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y)\geq0.$$

A: Well.  It's NOT true.  If $x = y = z > 0$ then $x^n(x-y)(x-z) + y^n(y-x)(y-z) + z^n(z-x)(z-y) = 0$.
So you can not prove it.
But maybe we can prove it is we stipulate that at least two of the terms are distinct.
Note that we can label the three variables whatever we want and by symmetry we might as well, and we can assume without loss of generality that $0 < x \le y \le z$.  (we can just relabel the variables so that is true.)
Case 1:  $x=y < z$
Then $x^n(x-z)(x-y) + y^n(y-z)(y-x) + z^n(z-y)(z-x) = 0+0 + z^n(z-y)^2 > 0$ as $(z-y)^2 > 0, z^n > 0$.
Case 2: $x < y=z$ 
Then $x^n(x-y)(x-z) + y^n(y-z)(y-x) + z^n(z-x)(z-y) = x^n(x-y)^2 + 0 + 0> 0$ as $x^n, (x-y)^2 > 0$.
Case 3: $x < y < z$.
Then $x^n(x-y)(x-z) > 0$ and $y^n(y-z)(y-x) < 0$ and $z^n(z-x)(z-y) > 0$.
So we need to prove that $x^n(y-x)(z-x) + z^n(z-x)(z-y) > y^n(z-y)(y-x)$. (Note: all terms are positive.)
That will be true if and only if $\frac {x^n}{z-y} + \frac {z^2}{y-x} > \frac {y^n}{z-x}$
So it if we can prove: $\frac {x^n}{z-y} + \frac {z^2}{y-x} > \frac {y^n}{z-x}$ we will be done.
$z-y < z-x$ and $y-x < z-x$ so 
$\frac {x^n}{z-y} + \frac {z^n}{y-x} >$
$ \frac {x^n}{z-x} + \frac{z^n}{z-x} > $
$\frac {z^n}{z-x} >$
$\frac {y^n}{z-x}$.
And that's it.
A: Hint:
Note that you have written down a slightly different version of Schur's inequality:


*

*$x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y) \color{red}{\geq} 0$ for $x,y,z \color{red}{\geq} 0$ and $t>0$
Equality holds iff $x=y=z$ or one of them is zero and the other two are equal.
A very simple proof of this inequality can also be found on the linked page.
A: Without loss of generality, assume $x\ge y\ge z\gt0$. Then,
$$
\begin{align}
\Delta
&=x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)\\
&=\underbrace{\ (x-y)\ \vphantom{\left[y^n\right]}}_{x\ge y}\underbrace{\left[x^n(x-z)-y^n(y-z)\right]}_{\substack{x^n\ge y^n\ge0\quad x-z\ge y-z\ge0\\x^n(x-z)\ge y^n(y-z)}}+\underbrace{z^n(x-z)(y-z)\vphantom{\left[y^n\right]}}_{z^n\ge0\ \ x-z\ge0\ \ y-z\ge0}\\
&\ge0
\end{align}
$$
If $x=y\gt z$, then $\Delta=z^n(y-z)^2\gt0$.
If $x\gt y=z$, then $\Delta=x^n(x-y)^2\gt0$.
Thus, $\Delta=0$ only if $x=y=z$.
