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Schur's Inequality

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Schur's Inequality is an inequality that holds for positive numbers. It is named for Issai Schur.


Contents

Theorem

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 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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Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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