# Difference between revisions of "Nesbitt's Inequality"

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with equality when all the variables are equal. | with equality when all the variables are equal. | ||

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

If <math> a_1, \ldots a_n </math> are positive and <math> \sum_{i=1}^{n}a_i = s </math>, then | If <math> a_1, \ldots a_n </math> are positive and <math> \sum_{i=1}^{n}a_i = s </math>, then |

## Revision as of 15:40, 29 April 2007

**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,

,

so

.

Since , weighted AM-HM gives us

.