Muirhead's inequality

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

a-mean
For any real vector


 * $$a=(a_1,\dots,a_n)$$

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


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

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

where ℓ is the number of distinct elements in a, and k1, ..., kℓ 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 $$a_1+\cdots+a_n=1$$. In the general case, one can consider instead $$[a]^{1/(a_1+\cdots+a_n)}$$, which is called a Muirhead mean.


 * 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
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:


 * $$a_1 \geq a_2 \geq \cdots \geq a_n$$


 * $$b_1 \geq b_2 \geq \cdots \geq b_n.$$

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



\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 $$b_1, \ldots, b_n$$ is said to majorize the sequence $$a_1, \ldots, a_n$$.

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 ($$\alpha_1, \ldots, \alpha_n$$) majorizes the other one.


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

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


 * $$\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}$$

Arithmetic-geometric mean inequality
Let
 * $$a_G = \left( \frac 1 n, \ldots , \frac 1 n \right)$$

and
 * $$a_A = ( 1, 0, 0, \ldots , 0 ).$$

We have

\begin{align} a_{A1} = 1 & > a_{G1} = \frac 1 n, \\ a_{A1} + a_{A2} = 1 & > 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
 * $$\frac 1 {n!} (x_1^1 \cdot x_2^0 \cdots x_n^0 + \cdots + x_1^0 \cdots x_n^1) (n-1)! \geq \frac 1 {n!} (x_1 \cdot \cdots \cdot x_n)^{1/n} n!$$

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:


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

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

Similarly, we can prove the inequality


 * $$x^3+y^3+z^3 \ge 3 x y z$$

by writing it using the symmetric-sum notation as


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

which is the same as


 * $$ 2 x^3 + 2 y^3 + 2 z^3 \ge 6 x y z. $$

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