- Schur's inequality
In
mathematics , Schur'sinequality , named afterIssai 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
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Inequality
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