Muirhead's Inequality

Muirhead's Average

For positive $\{x_i\}\;$ and non-negative $\{a_i\},\;$ $i=1,\ldots,n,\;$ we consider expression $F(x_1,x_2,\ldots,x_n)=x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n}\;$ and their cyclic sums that appear in the literature in various notations:

$\displaystyle\sum_{cycl}F(x_1,x_2,\ldots,x_n)=\sum !F(x_1,x_2,\ldots,x_n)=\sum_{\tau\in S_n}F(x_{\tau(1)},x_{\tau(2)},\ldots,x_{\tau(n)}),$

where $S_n\;$ is the group of all permutations of length $n.$

For example, if $\mathbf{a}=\{0,1,2\},\;$ then

$\sum !F(x_1,x_2,x_3)=x_2x_3^2+x_1x_2^2+x_3x_1^2+x_2x_1^2+x_1x_3^2+x_3x_2^2.$

Muirhead's Average for a sequence of exponents $\mathbf{a}=\{a_i\},\;$ $i=1,\ldots,n,\;$ is, by definition

$\displaystyle [\mathbf{a}]=\frac{1}{n!}\sum !F(x_1,x_2,\ldots,x_n)=\frac{1}{n!}\sum !x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n}.$

For example, for $\mathbf{a}=\{1,0,0,\ldots,0\},\;$ $\sum !x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n}=\sum!x_{1},\;$ where each $x_i\;$ appears $(n-1)!\;$ times such that Muirhead's Average for $\mathbf{a}=\{1,0,0,\ldots,0\}\;$ is just the arithmetic mean $\displaystyle\frac{1}{n}\sum !x_1=\frac{1}{n}(x_1+x_2+\ldots+x_n).$

For $\displaystyle\mathbf{a}=\{\frac{1}{n},\frac{1}{n},\frac{1}{n},\ldots,\frac{1}{n}\},\;$ $\sum !x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n}=\sum!\sqrt[n]{x_1x_2\ldots x_n},\;$ so that Muirhead's Average for $\displaystyle\{\frac{1}{n},\frac{1}{n},\frac{1}{n},\ldots,\frac{1}{n}\},\;$ is the geometric mean $\sqrt[n]{x_1x_2\ldots x_n}.\;$

Muirhead's Inequality

Muirhead's inequality provides a criterion for comparison of two Muirhead's averages of the same length. Muirhead's averages differ from one another by the sequence of exponents $\mathbf{a}.\;$ Thus it gives a procedure of comparing two such sequences.


Given two not increasing sequences $\mathbf{a}\;$ and $\mathbf{b},\;$ $\mathbf{a}\;$ is said to majorize $\mathbf{b}\;$ (written $\mathbf{a}\succ\mathbf{b})\;$ if


$\displaystyle\sum_{i=1}^{n}a_i= \sum_{i=1}^{n}b_i.$


$\displaystyle\sum_{i=1}^{k}a_i\ge \sum_{i=1}^{k}b_i.$

where $1\le k\lt n.$

Muirhead's Inequality

$\mathbf{a}\succ\mathbf{b}\;$ implies $[\mathbf{a}]\ge [\mathbf{b}].$

Equality is achieved only if either $\mathbf{a}=\mathbf{b},\;$ or all $x'\mbox{s}\;$ are equal.

For example, $\{1,0,0,\ldots,0\}\succ\displaystyle\{\frac{1}{n},\frac{1}{n},\frac{1}{n},\ldots,\frac{1}{n}\},\;$ which establishes the AM-GM inequality.


  1. Lau Chi Hin, Muirhead’s Inequality, Mathematicla Excalibur, Volume 11, Number 1, February - March 2006 (
  2. G.H. Hardy, J.E. Littlewood, G. Pólya, Inequalities, Cambridge Mathematical Library (2. ed.), Cambridge University Press, Section 2.18, Theorem 45.


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