# Difference between revisions of "Maclaurin's Inequality"

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Since the [[geometric mean]] of <math> 1/x_1, \ldots, 1/x_n </math> is 1, the inequality is true by [[AM-GM]]. | Since the [[geometric mean]] of <math> 1/x_1, \ldots, 1/x_n </math> is 1, the inequality is true by [[AM-GM]]. | ||

− | == | + | == See Also == |

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* [[Inequality]] | * [[Inequality]] | ||

* [[Newton's Inequality]] | * [[Newton's Inequality]] | ||

* [[Symmetric sum]] | * [[Symmetric sum]] | ||

− | [[Category: | + | [[Category:Algebra]] |

− | [[Category: | + | [[Category:Inequalities]] |

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## Latest revision as of 15:48, 29 December 2021

**Maclaurin's Inequality** is an inequality in symmetric polynomials. For notation and background, we refer to Newton's Inequality.

## Statement

For non-negative ,

,

with equality exactly when all the are equal.

## Proof

By the lemma from Newton's Inequality, it suffices to show that for any ,

.

Since this is a homogenous inequality, we may normalize so that . We then transform the inequality to

.

Since the geometric mean of is 1, the inequality is true by AM-GM.