# Difference between revisions of "Nesbitt's Inequality"

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− | '''Nesbitt's [[Inequality]]''' is a theorem which, although rarely cited, has many instructive proofs. It states that for positive <math> | + | '''Nesbitt's [[Inequality]]''' is a theorem which, although rarely cited, has many instructive proofs. It states that for positive <math>a, b, c </math>, |

<center> | <center> | ||

<math> | <math> | ||

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</math>, | </math>, | ||

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− | with equality when all the <math> | + | with equality when all the <math>a_i </math> are equal. |

== Proofs == | == Proofs == | ||

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=== By Rearrangement === | === By Rearrangement === | ||

− | Note that <math> | + | Note that <math>a,b,c </math> and <math> \frac{1}{b+c} = \frac{1}{a+b+c -a}</math>, <math> \frac{1}{c+a} = \frac{1}{a+b+c -b} </math>, <math> \frac{1}{a+b} = \frac{1}{a+b+c -c} </math> are sorted in the same order. Then by the [[rearrangement inequality]], |

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<math> | <math> | ||

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</math>. | </math>. | ||

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− | For equality to occur, since we changed <math>{} a \cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a} </math> to <math> b \cdot \frac{1}{b+c} + a \cdot \frac{1}{c+a} </math>, we must have <math> | + | For equality to occur, since we changed <math>{} a \cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a} </math> to <math> b \cdot \frac{1}{b+c} + a \cdot \frac{1}{c+a} </math>, we must have <math>a=b </math>, so by symmetry, all the variables must be equal. |

=== By Cauchy === | === By Cauchy === | ||

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</math>, | </math>, | ||

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− | as desired. Equality occurs when <math> | + | as desired. Equality occurs when <math>(b+c)^2 = (c+a)^2 = (a+b)^2 </math>, i.e., when <math>a=b=c </math>. |

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

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</math>. | </math>. | ||

</center> | </center> | ||

− | Setting <math> | + | Setting <math>x = b+c, y= c+a, z= a+b </math>, we expand the left side to obtain |

<center> | <center> | ||

<math> | <math> | ||

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</math>, | </math>, | ||

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− | which follows from <math> \frac{x}{y} + \frac{y}{x} \ge 2 </math>, etc., by [[AM-GM]], with equality when <math> | + | which follows from <math> \frac{x}{y} + \frac{y}{x} \ge 2 </math>, etc., by [[AM-GM]], with equality when <math>x=y=z </math>. |

==== By AM-HM ==== | ==== By AM-HM ==== | ||

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</math>. | </math>. | ||

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− | Setting <math> | + | Setting <math>x=b+c, y=c+a, z=a+b </math> yields the desired inequality. |

=== By Substitution === | === By Substitution === | ||

− | The numbers <math> | + | The numbers <math>x = \frac{a}{b+c}, y = \frac{b}{c+a}, z = \frac{c}{a+b} </math> satisfy the condition <math>xy + yz + zx + 2xyz = 1 </math>. Thus it is sufficient to prove that if any numbers <math>x,y,z </math> satisfy <math>xy + yz + zx + 2xyz = 1 </math>, then <math> x+y+z \ge \frac{3}{2} </math>. |

− | Suppose, on the contrary, that <math> x+y+z < \frac{3}{2} </math>. We then have <math> | + | Suppose, on the contrary, that <math> x+y+z < \frac{3}{2} </math>. We then have <math>xy + yz + zx \le \left( \frac{x+y+z}{3} \right)^2 < \frac{3}{4} </math>, and <math> 2xyz \le 2 \left( \frac{x+y+z}{3} \right)^3 < \frac{1}{4} </math>. Adding these inequalities yields <math>xy + yz + zx + 2xyz < 1 </math>, a contradiction. |

=== By Normalization and AM-HM === | === By Normalization and AM-HM === | ||

− | We may normalize so that <math> | + | We may normalize so that <math>a+b+c = 1 </math>. It is then sufficient to prove |

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<math> | <math> | ||

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=== By Weighted AM-HM === | === By Weighted AM-HM === | ||

− | We may normalize so that <math> | + | We may normalize so that <math>a+b+c =1 </math>. |

We first note that by the [[rearrangement inequality]], | We first note that by the [[rearrangement inequality]], | ||

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</center> | </center> | ||

− | Since <math> | + | Since <math>a+b+c = 1 </math>, weighted AM-HM gives us |

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<math> | <math> | ||

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</math>. | </math>. | ||

</center> | </center> | ||

+ | |||

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

+ | [[Category:Theorem]] |

## Revision as of 14:14, 26 October 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

.