en>Michael Hardy at 04:52, 13 July 2013

2013-07-13T04:52:37Z

← Older revision		Revision as of 06:52, 13 July 2013
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	~~Nice to satisfy you~~, ~~I am Marvella Shryock. One~~ of the ~~very best things in~~ the ~~world~~ for ~~me is to do aerobics~~ and I'~~ve been doing it for quite a whilst~~. ~~Puerto Rico~~ is ~~where he~~'s ~~usually been residing but she requirements to move because~~ of ~~her family members. My working day job~~ is a ~~meter reader~~.<br><br>~~My website~~ [~~http~~://~~Computerrepairredlands~~.~~com~~/~~solid~~-~~advice~~-~~terms~~-~~candida~~-~~albicans~~/ ~~Computerrepairredlands~~.~~com~~]		In [[mathematics]], '''Schur's [[inequality (mathematics)\|inequality]]''', named after [[Issai Schur]],
			establishes that for all [[Nonnegative number\|non-negative]] [[real number]]s
			''x'', ''y'', ''z'' and a [[positive number]] ''t'',

			:<math>x^t (x-y)(x-z) + y^t (y-z)(y-x) + z^t (z-x)(z-y) \ge 0</math>

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

			== Proof ==
			Since the inequality is symmetric in <math>x,y,z</math> we may assume without loss of generality that <math> x \geq y \geq z</math>. Then the inequality

			: <math>(x-y)[x^t(x-z)-y^t(y-z)]+z^t(x-z)(y-z) \geq 0\,</math>

			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 [[Order isomorphic\|similarly sorted]], then the following inequality holds:

			:<math>a (x-y)(x-z) + b (y-z)(y-x) + c (z-x)(z-y) \ge 0.</math>

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

			Consider <math>a,b,c,x,y,z \in \mathbb{R}</math>, where <math>a \geq b \geq c</math>, and either <math>x \geq y \geq z</math> or <math>z \geq y \geq x</math>. Let <math>k \in \mathbb{Z}^{+}</math>, and let <math>f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}</math> be either [[convex function\|convex]] or [[monotonic]]. Then,
			: <math>{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}.\,</math>
			The standard form of Schur's is the case of this inequality where ''x'' = ''a'', ''y'' = ''b'', ''z'' = ''c'', ''k'' = 1, ƒ(''m'') = ''m''<sup>''r''</sup>.<ref>Vornicu, Valentin; ''Olimpiada de Matematica... de la provocare la experienta''; GIL Publishing House; Zalau, Romania.</ref>

			==Notes==
			{{reflist}}

			[[Category:Inequalities]]
			[[Category:Articles containing proofs]]

en>David Eppstein: Reverted edits by 70.247.166.5 (talk) to last version by Citation bot 1

2012-06-23T18:11:24Z

Reverted edits by 70.247.166.5 (talk) to last version by Citation bot 1

New page

Nice to satisfy you, I am Marvella Shryock. One of the very best things in the world for me is to do aerobics and I've been doing it for quite a whilst. Puerto Rico is where he's usually been residing but she requirements to move because of her family members. My working day job is a meter reader.<br><br>My website [http://Computerrepairredlands.com/solid-advice-terms-candida-albicans/ Computerrepairredlands.com]

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en>Michael Hardy at 04:52, 13 July 2013

en>David Eppstein: Reverted edits by 70.247.166.5 (talk) to last version by Citation bot 1