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Difference between revisions of "Schur's Inequality"

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<math>{a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0}</math>
 
<math>{a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0}</math>
  
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The four equality cases occur when <math>a=b=c</math>, or when two of <math>a,b,c</math> are equal and the third is <math>{0}</math>.
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The four [[equality condition | equality cases]] occur when <math>a=b=c</math> or when two of <math>a,b,c</math> are equal and the third is <math>{0}</math>.
  
  

Revision as of 08:24, 28 July 2006

Schur's Inequality states that for all non-negative $a,b,c \in \mathbb{R}$ and $r>0$:

${a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0}$

The four equality cases occur when $a=b=c$ or when two of $a,b,c$ are equal and the third is ${0}$.


Common Cases

The $r=1$ case yields the well-known inequality:$a^3+b^3+c^3+3abc \geq a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b$

When $r=2$, an equivalent form is: $a^4+b^4+c^4+abc(a+b+c) \geq a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b$


Proof

WLOG, let ${a \geq b \geq c}$. Note that $\displaystyle a^r(a-b)(a-c)+b^r(b-a)(b-c) = a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))$. Clearly, $a^r \geq b^r \geq 0$, and $a-c \geq b-c \geq 0$. Thus, $(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \rightarrow a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0$. However, $c^r(c-a)(c-b) \geq 0$, and thus the proof is complete.


Generalized Form

It has been shown by Valentin Vornicu that a more general form of Schur's Inequality exists. Consider $a,b,c,x,y,z \in \mathbb{R}$, where ${a \geq b \geq c}$, and either $x \geq y \geq z$ or $z \geq y \geq x$. Let $k \in \mathbb{Z}^{+}$, and let $f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}$ be either convex or monotonic. Then,

${f(x)(a-b)^k(a-c)^k+f(y)(b-a)^k(b-c)^k+f(z)(c-a)^k(c-b)^k \geq 0}$

The standard form of Schur's is the case of this inequality where $x=a,\ y=b,\ z=c,\ k=1,\ f(m)=m^r$.


References

  • Vornicu, Valentin; Olimpiada de Matematica... de la provocare la experienta; GIL Publishing House; Zalau, Romania.

See also

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