Difference between revisions of "Schur's Inequality"

Revision as of 08:35, 23 April 2021

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

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(Without loss of Generality), 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, $(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0$ . 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

Mildorf, Thomas; Olympiad Inequalities; January 20, 2006; <http://artofproblemsolving.com/articles/files/MildorfInequalities.pdf>

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

@@ Line 2: / Line 2: @@
-== Theorem ==
+== THEOREM ==
 Schur's inequality states that for all non-negative <math>a,b,c \in \mathbb{R}</math> and <math>r>0</math>:
@@ Line 20: / Line 20: @@
 === Proof ===
-[[WLOG]], let <math>{a \geq b \geq c}</math>.  Note that <math>a^r(a-b)(a-c)+b^r(b-a)(b-c)</math> <math>= a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))</math>.  Clearly, <math>a^r \geq b^r \geq 0</math>, and <math>a-c \geq b-c \geq 0</math>.  Thus, <math>(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0</math>.  However, <math>c^r(c-a)(c-b) \geq 0</math>, and thus the proof is complete.
+[[WLOG(Without loss of Generality)]], let <math>{a \geq b \geq c}</math>.  Note that <math>a^r(a-b)(a-c)+b^r(b-a)(b-c)</math> <math>= a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))</math>.  Clearly, <math>a^r \geq b^r \geq 0</math>, and <math>a-c \geq b-c \geq 0</math>.  Thus, <math>(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0</math>.  However, <math>c^r(c-a)(c-b) \geq 0</math>, and thus the proof is complete.
 === Generalized Form ===

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