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>0'',

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

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

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) ^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:

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

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

Consider $a,b,c,x,y,z \in \mathbb$, where $a \geq b \geq c$, and either $x \geq y \geq z$ or $z \geq y \geq x$. Let $k \in \mathbb^$, and let $f:\mathbb \rightarrow \mathbb_^$ be either convex or monotonic. Then,
: $.$
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 $x \geq y \geq z \geq v$ with and the positive real number ''t'' are such that ''x'' + ''v'' ≥ ''y'' + ''z'' then

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

Notes

{{reflist

Inequalities
Articles containing proofs
Issai Schur