- Schur's inequality
In

mathematics ,**Schur's**, named afterinequality Issai Schur ,establishes that for all non-negativereal number s"x", "y", "z" and apositive 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)" aresimilarly 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 ,Romania n mathematicianValentin 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

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