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.