# Schur's inequality

In mathematics, Schur's inequality, named after Issai Schur, establishes that for all non-negative real numbers x, y, z and 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 inequality is non-negative. This rearranges to Schur's inequality.

## Extensions

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) = mr.

Another possible extension states that if the non-negative real numbers $x\geq y\geq z\geq v$  with and the positive real number t are such that x + v ≥ y + z then

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