# Difference between revisions of "Nesbitt's Inequality"

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By Cauchy, <cmath>\sum_{\text{cyc}}\frac{a^2}{ab + ac} \ge \frac{(a + b + c)^2}{2(ab + ac + bc)} = \frac{a^2 + b^2 + c^2 + 2(ab + ac + bc)}{2(ab + ac + bc)}</cmath> <cmath> = 1 + \frac{a^2 + b^2 + c^2}{2(ab + ac + bc)} \ge \frac{3}{2}</cmath> since <math>a^2 + b^2 + c^2 \ge ab + ac + bc</math> by Muirhead as <math>[1, 1, 0]\prec [2, 0, 0]</math> | By Cauchy, <cmath>\sum_{\text{cyc}}\frac{a^2}{ab + ac} \ge \frac{(a + b + c)^2}{2(ab + ac + bc)} = \frac{a^2 + b^2 + c^2 + 2(ab + ac + bc)}{2(ab + ac + bc)}</cmath> <cmath> = 1 + \frac{a^2 + b^2 + c^2}{2(ab + ac + bc)} \ge \frac{3}{2}</cmath> since <math>a^2 + b^2 + c^2 \ge ab + ac + bc</math> by Muirhead as <math>[1, 1, 0]\prec [2, 0, 0]</math> | ||

+ | |||

+ | ===Another Interesting Method=== | ||

+ | |||

+ | Let <cmath>S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}</cmath> | ||

+ | And <cmath>M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}</cmath> | ||

+ | And <cmath>N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}</cmath> | ||

+ | Now, we get <cmath>M+N=3</cmath> | ||

+ | Also by AM-GM; <cmath>M+S\geq 3</cmath> and <cmath>N+S\geq 3</cmath> | ||

+ | <cmath>\implies M+N+2S\geq 6</cmath> | ||

+ | <cmath>\implies 2S\geq 3</cmath> | ||

+ | <cmath>\implies S\geq \frac{3}{2}</cmath> | ||

[[Category:Inequality]] | [[Category:Inequality]] | ||

[[Category:Theorems]] | [[Category:Theorems]] |

## Revision as of 21:59, 28 May 2015

**Nesbitt's Inequality** is a theorem which, although rarely cited, has many instructive proofs. It states that for positive ,

,

with equality when all the variables are equal.

All of the proofs below generalize to proof the following more general inequality.

If are positive and , then

,

or equivalently

,

with equality when all the are equal.

## Contents

## Proofs

### By Rearrangement

Note that and , , are sorted in the same order. Then by the rearrangement inequality,

.

For equality to occur, since we changed to , we must have , so by symmetry, all the variables must be equal.

### By Cauchy

By the Cauchy-Schwarz Inequality, we have

,

or

,

as desired. Equality occurs when , i.e., when .

We also present three closely related variations of this proof, which illustrate how AM-HM is related to AM-GM and Cauchy.

#### By AM-GM

By applying AM-GM twice, we have

,

which yields the desired inequality.

#### By Expansion and AM-GM

We consider the equivalent inequality

.

Setting , we expand the left side to obtain

,

which follows from , etc., by AM-GM, with equality when .

#### By AM-HM

The AM-HM inequality for three variables,

,

is equivalent to

.

Setting yields the desired inequality.

### By Substitution

The numbers satisfy the condition . Thus it is sufficient to prove that if any numbers satisfy , then .

Suppose, on the contrary, that . We then have , and . Adding these inequalities yields , a contradiction.

### By Normalization and AM-HM

We may normalize so that . It is then sufficient to prove

,

which follows from AM-HM.

### By Weighted AM-HM

We may normalize so that .

We first note that by the rearrangement inequality or the fact that ,

,

so

.

Since , weighted AM-HM gives us

.

### By Muirhead's and Cauchy

By Cauchy, since by Muirhead as

### Another Interesting Method

Let And And Now, we get Also by AM-GM; and