Schur's inequality

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

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

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.

1 Proof
2 Extension
3 Notes
4 See also

[edit] 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 equation is non-negative. This rearranges to Schur's inequality.

[edit] Extension

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 a further generalized form of Schur's inequality exists:

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,