Schur's Inequality

Theorem
Let $x, y, z \in \R_+$ be non-negative real numbers.

Let $t \in \R, t > 0$ be a (strictly) positive real number.

Then:
 * $x^t \left({x-y}\right) \left({x-z}\right) + y^t \left({y-z}\right) \left({y-x}\right)) + z^t \left({z-x}\right) \left({z-y}\right) \ge 0$

The equality holds iff either:
 * $x = y = z$;
 * Two of them are equal and the other is zero.

When $t$ is a positive even integer, the inequality holds for all real numbers $x, y, z$.

Proof
We note that the inequality, as stated, is symmetrical in $x, y$ and $z$.

So WLOG we can assume that $ x \ge y \ge z$.

Consider the expression:
 * $\left({x-y}\right) \left({x^t \left({x-z}\right) - y^t \left({y-z}\right)}\right) + z^t \left({x-z}\right) \left({y-z}\right)$

We see that every term in the above is non-negative. So, directly:
 * $(1) \qquad \left({x-y}\right) \left({x^t \left({x-z}\right) - y^t \left({y-z}\right)}\right) + z^t \left({x-z}\right) \left({y-z}\right) \ge 0$

If $x = y = z = 0$, it clearly evaluates to $0$.

Inspection on a case-by-case basis provides evidence for the other conditions for equality.

$(1)$ can then be rearranged to Schur's inequality.