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.

When t=1, the following well-known special case can be derived:

x^3 + y^3 + z^3 + 3xyz \geq xy(x+y) + xz(x+z) + yz(y+z)

Proof[edit]

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[edit]

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} \rightarrow \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) = mr.[1]

Notes[edit]

  1. ^ Vornicu, Valentin; Olimpiada de Matematica... de la provocare la experienta; GIL Publishing House; Zalau, Romania.