Schur’s inequality


If a, b, and c are non-negative real numbers and k1 is real, then the following inequality holds:

ak(a-b)(a-c)+bk(b-c)(b-a)+ck(c-a)(c-b)0
Proof.

We can assume without loss of generality that cba via a permutationMathworldPlanetmath of the variables (as both sides are symmetricPlanetmathPlanetmath in those variables). Then collecting terms, we wish to show that

(a-b)(ak(a-c)-bk(b-c))+ck(a-c)(b-c)0

which is clearly true as every term on the left is positive.∎

There are a couple of special cases worth noting:

  • Taking k=1, we get the well-known

    a3+b3+c3+3abcab(a+b)+ac(a+c)+bc(b+c)
  • If c=0, we get (a-b)(ak+1-bk+1)0.

  • If b=c=0, we get ak+20.

  • If b=c, we get ak(a-c)20.

Title Schur’s inequality
Canonical name SchursInequality
Date of creation 2013-03-22 13:19:30
Last modified on 2013-03-22 13:19:30
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 11
Author rspuzio (6075)
Entry type Theorem
Classification msc 26D15