Schur's Inequality

Since the inequality is symmetric in ${\displaystyle x,y,z}$ we may assume without loss of generality that ${\displaystyle x\geq y\geq z}$. Then the inequality

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

clearly holds, since every term on the left-hand side of the inequality is non-negative. This rearranges to Schur's inequality.

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:

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

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

Consider ${\displaystyle a,b,c,x,y,z\in \mathbb {R} }$, where ${\displaystyle a\geq b\geq c}$, and either ${\displaystyle x\geq y\geq z}$ or ${\displaystyle z\geq y\geq x}$. Let ${\displaystyle k\in \mathbb {Z} ^{+}}$, and let ${\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} _{0}^{+}}$ be either convex or monotonic. Then,

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

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.[1]

Another possible extension states that if the non-negative real numbers ${\displaystyle x\geq y\geq z\geq v}$ with and the positive real number t are such that x + v ≥ y + z then[2]

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