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,

$x^{t}(x-y)(x-z)+y^{t}(y-z)(y-x)+z^{t}(z-x)(z-y)\geq 0$

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

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:

$a(x-y)(x-z)+b(y-z)(y-x)+c(z-x)(z-y)\geq 0.$

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) = m^r.