Schur's inequality

In mathematics, Schur's inequality, named after Issai Schur, establishes that for all non-negative real numbers x, y, z, and t>0,


 * $$x^t (x-y)(x-z) + y^t (y-z)(y-x) + z^t (z-x)(z-y) \ge 0$$

with equality if and only if x = y = z or two of them are equal and the other is zero. When t is an even positive integer, the inequality holds for all real numbers x, y and z.

When $$t=1$$, the following well-known special case can be derived:
 * $$x^3 + y^3 + z^3 + 3xyz \geq xy(x+y) + xz(x+z) + yz(y+z)$$

Proof
Since the inequality is symmetric in $$x,y,z$$ we may assume without loss of generality that $$ x \geq y \geq z$$. Then the inequality


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

clearly holds, since every term on the left-hand side of the inequality is non-negative. This rearranges to Schur's inequality.

Extensions
A generalization of Schur's inequality is the following: Suppose a,b,c are positive real numbers. If the triples (a,b,c) and (x,y,z) are similarly sorted, then the following inequality holds:


 * $$a (x-y)(x-z) + b (y-z)(y-x) + c (z-x)(z-y) \ge 0.$$

In 2007, Romanian mathematician Valentin Vornicu showed that a yet further generalized form of Schur's inequality holds:

Consider $$a,b,c,x,y,z \in \mathbb{R}$$, where $$a \geq b \geq c$$, and either $$x \geq y \geq z$$ or $$z \geq y \geq x$$. Let $$k \in \mathbb{Z}^{+}$$, and let $$f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}$$ be either convex or monotonic. Then,
 * $${f(x)(a-b)^k(a-c)^k+f(y)(b-a)^k(b-c)^k+f(z)(c-a)^k(c-b)^k \geq 0}.$$

The standard form of Schur's is the case of this inequality where x = a, y = b, z = c, k = 1, ƒ(m) = mr.

Another possible extension states that if the non-negative real numbers $$ x \geq y \geq z \geq v $$ with and the positive real number t are such that x + v ≥ y + z then


 * $$x^t (x-y)(x-z)(x-v) + y^t (y-x)(y-z)(y-v) + z^t (z-x)(z-y)(z-v) + v^t (v-x)(v-y)(v-z) \ge 0. $$