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Schur's inequality

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This article is about the algebraic inequality in 3 variables. For the integral inequality, see Schur test.
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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 ) ≥ 0 {\displaystyle 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 {\displaystyle t=1} , the following well-known special case can be derived:

x 3 + y 3 + z 3 + 3 x y z ≥ x y ( x + y ) + x z ( x + z ) + y z ( y + z ) {\displaystyle 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 {\displaystyle x,y,z} we may assume without loss of generality that x ≥ y ≥ z {\displaystyle x\geq y\geq z} . Then the inequality

( x − y ) [ x t ( x − z ) − y t ( y − z ) ] + z t ( x − z ) ( y − z ) ≥ 0 {\displaystyle (x-y)+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 ) ≥ 0. {\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 a , b , c , x , y , z ∈ R {\displaystyle a,b,c,x,y,z\in \mathbb {R} } , where a ≥ b ≥ c {\displaystyle a\geq b\geq c} , and either x ≥ y ≥ z {\displaystyle x\geq y\geq z} or z ≥ y ≥ x {\displaystyle z\geq y\geq x} . Let k ∈ Z + {\displaystyle k\in \mathbb {Z} ^{+}} , and let f : R → R 0 + {\displaystyle 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 ≥ 0 . {\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.

Another possible extension states that if the non-negative real numbers x ≥ y ≥ z ≥ v {\displaystyle 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 ) ≥ 0. {\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.}

Notes

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