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Schur's inequality

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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 a positive number t,

Contents

x t ( x y ) ( x z ) + y t ( y z ) ( y x ) + z t ( z x ) ( z y ) 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 + 3 x y z x y ( x + y ) + x z ( x + z ) + y z ( y + z )

Proof

Since the inequality is symmetric in x , y , z we may assume without loss of generality that x y z . Then the inequality

( x y ) [ x t ( x z ) y t ( y z ) ] + z t ( x z ) ( y z ) 0

clearly holds, since every term on the left-hand side of the equation 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.

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 , where a b c , and either x y z or z y x . Let k Z + , 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 0 .

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

Another possible extension states that if the non-negative real numbers x y z 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.

References

Schur's inequality Wikipedia