In mathematics, Muirhead's inequality, named after Robert Franklin Muirhead, also known as the "bunching" method, generalizes the inequality of arithmetic and geometric means.

## Preliminary definitions

### a-mean

For any real vector

$\displaystyle{ a=(a_1,\dots,a_n) }$

define the "a-mean" [a] of positive real numbers x1, ..., xn by

$\displaystyle{ [a]=\frac{1}{n!}\sum_\sigma x_{\sigma_1}^{a_1}\cdots x_{\sigma_n}^{a_n}, }$

where the sum extends over all permutations σ of { 1, ..., n }.

When the elements of a are nonnegative integers, the a-mean can be equivalently defined via the monomial symmetric polynomial $\displaystyle{ m_a(x_1,\dots,x_n) }$ as

$\displaystyle{ [a] = \frac{k_1!\cdots k_l!}{n!} m_a(x_1,\dots,x_n), }$

where ℓ is the number of distinct elements in a, and k1, ..., k are their multiplicities.

Notice that the a-mean as defined above only has the usual properties of a mean (e.g., if the mean of equal numbers is equal to them) if $\displaystyle{ a_1+\cdots+a_n=1 }$. In the general case, one can consider instead $\displaystyle{ [a]^{1/(a_1+\cdots+a_n)} }$, which is called a Muirhead mean.[1]

Examples
• For a = (1, 0, ..., 0), the a-mean is just the ordinary arithmetic mean of x1, ..., xn.
• For a = (1/n, ..., 1/n), the a-mean is the geometric mean of x1, ..., xn.
• For a = (x, 1 − x), the a-mean is the Heinz mean.
• The Muirhead mean for a = (−1, 0, ..., 0) is the harmonic mean.

### Doubly stochastic matrices

Main page: Doubly stochastic matrix

An n × n matrix P is doubly stochastic precisely if both P and its transpose PT are stochastic matrices. A stochastic matrix is a square matrix of nonnegative real entries in which the sum of the entries in each column is 1. Thus, a doubly stochastic matrix is a square matrix of nonnegative real entries in which the sum of the entries in each row and the sum of the entries in each column is 1.

## Statement

Muirhead's inequality states that [a] ≤ [b] for all x such that xi > 0 for every i ∈ { 1, ..., n } if and only if there is some doubly stochastic matrix P for which a = Pb.

Furthermore, in that case we have [a] = [b] if and only if a = b or all xi are equal.

The latter condition can be expressed in several equivalent ways; one of them is given below.

The proof makes use of the fact that every doubly stochastic matrix is a weighted average of permutation matrices (Birkhoff-von Neumann theorem).

### Another equivalent condition

Because of the symmetry of the sum, no generality is lost by sorting the exponents into decreasing order:

$\displaystyle{ a_1 \geq a_2 \geq \cdots \geq a_n }$
$\displaystyle{ b_1 \geq b_2 \geq \cdots \geq b_n. }$

Then the existence of a doubly stochastic matrix P such that a = Pb is equivalent to the following system of inequalities:

\displaystyle{ \begin{align} a_1 & \leq b_1 \\ a_1+a_2 & \leq b_1+b_2 \\ a_1+a_2+a_3 & \leq b_1+b_2+b_3 \\ & \,\,\, \vdots \\ a_1+\cdots +a_{n-1} & \leq b_1+\cdots+b_{n-1} \\ a_1+\cdots +a_n & = b_1+\cdots+b_n. \end{align} }

(The last one is an equality; the others are weak inequalities.)

The sequence $\displaystyle{ b_1, \ldots, b_n }$ is said to majorize the sequence $\displaystyle{ a_1, \ldots, a_n }$.

## Symmetric sum notation

It is convenient to use a special notation for the sums. A success in reducing an inequality in this form means that the only condition for testing it is to verify whether one exponent sequence ($\displaystyle{ \alpha_1, \ldots, \alpha_n }$) majorizes the other one.

$\displaystyle{ \sum_\text{sym} x_1^{\alpha_1} \cdots x_n^{\alpha_n} }$

This notation requires developing every permutation, developing an expression made of n! monomials, for instance:

\displaystyle{ \begin{align} \sum_\text{sym} x^3 y^2 z^0 &= x^3 y^2 z^0 + x^3 z^2 y^0 + y^3 x^2 z^0 + y^3 z^2 x^0 + z^3 x^2 y^0 + z^3 y^2 x^0 \\ &= x^3 y^2 + x^3 z^2 + y^3 x^2 + y^3 z^2 + z^3 x^2 + z^3 y^2 \end{align} }

## Examples

### Arithmetic-geometric mean inequality

Main page: Inequality of arithmetic and geometric means

Let

$\displaystyle{ a_G = \left( \frac 1 n , \ldots , \frac 1 n \right) }$

and

$\displaystyle{ a_A = ( 1 , 0, 0, \ldots , 0 ). }$

We have

\displaystyle{ \begin{align} a_{A1} = 1 & \gt a_{G1} = \frac 1 n, \\ a_{A1} + a_{A2} = 1 & \gt a_{G1} + a_{G2} = \frac 2 n, \\ & \,\,\, \vdots \\ a_{A1} + \cdots + a_{An} & = a_{G1} + \cdots + a_{Gn} = 1. \end{align} }

Then

[aA] ≥ [aG],

which is

$\displaystyle{ \frac 1 {n!} (x_1^1 \cdot x_2^0 \cdots x_n^0 + \cdots + x_1^0 \cdots x_n^1) (n-1)! \geq \frac 1 {n!} (x_1 \cdot \cdots \cdot x_n)^{1/n} n! }$

yielding the inequality.

### Other examples

We seek to prove that x2 + y2 ≥ 2xy by using bunching (Muirhead's inequality). We transform it in the symmetric-sum notation:

$\displaystyle{ \sum_ \mathrm{sym} x^2 y^0 \ge \sum_\mathrm{sym} x^1 y^1. }$

The sequence (2, 0) majorizes the sequence (1, 1), thus the inequality holds by bunching.

Similarly, we can prove the inequality

$\displaystyle{ x^3+y^3+z^3 \ge 3 x y z }$

by writing it using the symmetric-sum notation as

$\displaystyle{ \sum_ \mathrm{sym} x^3 y^0 z^0 \ge \sum_\mathrm{sym} x^1 y^1 z^1, }$

which is the same as

$\displaystyle{ 2 x^3 + 2 y^3 + 2 z^3 \ge 6 x y z. }$

Since the sequence (3, 0, 0) majorizes the sequence (1, 1, 1), the inequality holds by bunching.