# Difference between revisions of "Muirhead's Inequality"

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− | '''Muirhead's Inequality''' states that if a sequence <math>A</math> [[Majorization|majorizes]] a sequence <math>B</math>, then given a set of positive | + | '''Muirhead's Inequality''' states that if a sequence <math>A</math> [[Majorization|majorizes]] a sequence <math>B</math>, then given a set of positive reals <math>x_1,x_2,\cdots,x_n</math>: |

<math>\sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}</math> | <math>\sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}</math> |

## Revision as of 22:37, 22 April 2008

**Muirhead's Inequality** states that if a sequence majorizes a sequence , then given a set of positive reals :

## Example

The inequality is easier to understand given an example. Since the sequence majorizes , Muirhead's inequality states that for any positive ,

.

## Usage on Olympiad Problems

A common bruteforce technique with inequalities is to clear denominators, multiply everything out, and apply Muirhead's or Schur's. However, it is worth noting that any inequality that can be proved directly with Muirhead can also be proved using the Arithmetic Mean-Geometric Mean inequality. In fact, IMO gold medalist Thomas Mildorf says it is unwise to use Muirhead in an Olympiad solution; one should use an application of AM-GM instead. Thus, it is suggested that Muirhead be used only to verify that an inequality *can* be proved with AM-GM before demonstrating the full AM-GM proof.

As an example, the above inequality can be proved using AM-GM as follows:

Adding these,

.

Multiplying both sides by (as both and are positive),

as desired.