Schur’s Inequality

For $a,b,c,r \in \mathbb{R}_0^+$,

$a^{r}(a-b)(a-c) + b^{r}(b-a)(b-c) + c^{r}(c-a)(c-b) \ge 0$

Proof

Since the inequality is symmetric we may assume without loss of generality that $x \geq y \geq z$. Then the inequality is equivalent to –

$(x-y)[x^t(x-z)-y^t(y-z)]+z^t(x-z)(y-z) \geq 0$

which is true from the assumption.