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Schur's inequality (Theorem)

If $ a$, $ b$, and $ c$ are non-negative real numbers and $ k\geq 1$ is real, then the following inequality holds:

$\displaystyle a^k(a-b)(a-c)+b^k(b-c)(b-a)+c^k(c-a)(c-b)\geq 0 $
Proof. We can assume without loss of generality that $ c\leq b\leq a$ via a permutation of the variables (as both sides are symmetric in those variables). Then collecting terms, we wish to show that
$\displaystyle (a-b)\left(a^k(a-c)-b^k(b-c)\right)+c^k(a-c)(b-c)\geq 0$    

which is clearly true as every term on the left is positive. $ \qedsymbol$

There are a couple of special cases worth noting:

  • Taking $ k=1$, we get the well-known
    $\displaystyle a^3 + b^3 + c^3 + 3abc \geq ab(a+b) + ac(a+c) + bc(b+c) $
  • If $ c=0$, we get $ (a-b)(a^{k+1}-b^{k+1})\geq0$.
  • If $ b=c=0$, we get $ a^{k+2}\geq0$.
  • If $ b=c$, we get $ a^{k}(a-c)^2\geq 0$.



"Schur's inequality" is owned by rspuzio. [ full author list (3) | owner history (2) ]
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Keywords:  sum, inequality, Schur
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Cross-references: positive, terms, symmetric, sides, variables, permutation, without loss of generality, inequality, real numbers
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This is version 8 of Schur's inequality, born on 2002-12-26, modified 2007-07-27.
Object id is 3836, canonical name is SchursInequality.
Accessed 6499 times total.

Classification:
AMS MSC: 26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals)
Pending Errata and Addenda
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