Schur's inequality

Template:Short description Template:About Template:More citations needed In mathematics, Schur's inequality, named after Issai Schur, establishes that for all non-negative real numbers x, y, z, and t,

$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\ge 0$

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:

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

Proof

Since the inequality is symmetric in $x,y,z$ we may assume without loss of generality that $x\ge y\ge z$. If $t\ge 0$ the inequality

$(x-y)[x^t(x-z)-y^t(y-z)]+z^t(x-z)(y-z)\ge 0$

clearly holds, since every term on the left-hand side of the inequality is non-negative. This rearranges to Schur's inequality. Similarly, when $t<0$,

$(y-z)[z^t(x-z)-y^t(x-y)]+x^t(x-y)(x-z)\ge 0$

This is also clearly true because the left-hand side consists of every term being non-negative.

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:

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

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

Consider $a,b,c,x,y,z\in ℝ$, where $a\ge b\ge c$, and either $x\ge y\ge z$ or $z\ge y\ge x$. Let $k\in {\mathbb{Z}}^{+}$, and let $f:ℝ\to {ℝ}_0^{+}$ be either convex or monotonic. Then,

$f(x)(a-b{)}^k(a-c{)}^k+f(y)(b-a{)}^k(b-c{)}^k+f(z)(c-a{)}^k(c-b{)}^k\ge 0.$

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.

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

$x^t(x-y)(x-z)(x-v)+y^t(y-x)(y-z)(y-v)+z^t(z-x)(z-y)(z-v)+v^t(v-x)(v-y)(v-z)\ge 0.$