Schur's inequality explained

In mathematics, Schur's inequality, named after Issai Schur,establishes that for all non-negative real numbersx, y, z, and t>0,

xt(x-y)(x-z)+yt(y-z)(y-x)+zt(z-x)(z-y)\ge0

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:

x3+y3+z3+3xyz\geqxy(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\geqy\geqz

. Then the inequality

(x-y)[xt(x-z)-yt(y-z)]+zt(x-z)(y-z)\geq0

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)\ge0.

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\inR

, where

a\geqb\geqc

, and either

x\geqy\geqz

or

z\geqy\geqx

. Let

k\inZ+

, and let

f:R

+
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\geq0}.

The standard form of Schur's is the case of this inequality where x = a, y = b, z = c, k = 1, ƒ(m) = mr.[1]

Another possible extension states that if the non-negative real numbers

x\geqy\geqz\geqv

with and the positive real number t are such that x + v ≥ y + z then[2]

xt(x-y)(x-z)(x-v)+yt(y-x)(y-z)(y-v)+zt(z-x)(z-y)(z-v)+vt(v-x)(v-y)(v-z)\ge0.

Notes and References

  1. Vornicu, Valentin; Olimpiada de Matematica... de la provocare la experienta; GIL Publishing House; Zalau, Romania.
  2. Finta. Béla. A Schur Type Inequality for Five Variables. Procedia Technology. 2015. 19. 799–801. 10.1016/j.protcy.2015.02.114. free.