Schur's inequality

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".

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.

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 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 $f:mathbb\{R\}\; ightarrow\; mathbb\{R\}\_\{0\}^\{+\}$ 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") = "m"^"r". [Vornicu, Valentin; "Olimpiada de Matematica... de la provocare la experienta"; GIL Publishing House; Zalau, Romania.]

Notes

ee also

*Inequality