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.

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, ..., x_n 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_l 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, ..., x_n.
• For a = (1/n, ..., 1/n), the a-mean is the geometric mean of x₁, ..., x_n.
• 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 P^T 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 x_i > 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 x_i 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:

${\displaystyle {\begin{aligned}a_{1}&\leq b_{1}\\a_{1}+a_{2}&\leq b_{1}+b_{2}\\a_{1}+a_{2}+a_{3}&\leq b_{1}+b_{2}+b_{3}\\&\,\,\,\vdots \\a_{1}+\cdots +a_{n-1}&\leq b_{1}+\cdots +b_{n-1}\\a_{1}+\cdots +a_{n}&=b_{1}+\cdots +b_{n}.\end{aligned}}}$

(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:

${\displaystyle {\begin{aligned}\sum _{\text{sym}}x^{3}y^{2}z^{0}&=x^{3}y^{2}z^{0}+x^{3}z^{2}y^{0}+y^{3}x^{2}z^{0}+y^{3}z^{2}x^{0}+z^{3}x^{2}y^{0}+z^{3}y^{2}x^{0}\\&=x^{3}y^{2}+x^{3}z^{2}+y^{3}x^{2}+y^{3}z^{2}+z^{3}x^{2}+z^{3}y^{2}\end{aligned}}}$

Examples

Arithmetic-geometric mean inequality

Let

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

and

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

We have

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

Then

[a_A] ≥ [a_G],

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:

${\displaystyle \sum _{\mathrm {sym} }x^{2}y^{0}\geq \sum _{\mathrm {sym} }x^{1}y^{1}.}$

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

Similarly, we can prove the inequality

${\displaystyle x^{3}+y^{3}+z^{3}\geq 3xyz}$

대칭-섬 표기법을 사용하여 다음과 같이 작성함

${\displaystyle \sum \sum \sum \sum \sum{3}y^{0}z^{0}\geq \sum _{\mathrm {sum}{}x^{1}y^{1}z^{1},}}}$

어느 것이 와 같은가.

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

순서(3, 0, 0)가 순서(1, 1, 1)를 전공하기 때문에 부등식은 뭉쳐 지탱한다.

참고 항목

• 산술 및 기하학적 평균의 부등식
• 이중 확률 매트릭스
• 단항 대칭 다항식

메모들

참조

• 1998년 지안 카를로 로타(Gian-Carlo Rota)의 강연에 근거한 John N. Guidi의 조합 이론, 2002년 MIT 복사 기술 센터.
• Kiran Kedlaya, A < B (B보다 적은 A), 불평등 해소에 대한 안내서
• 플래닛매트릭스에서 뮤어헤드의 정리.
• 하디, G.H.; 리틀우드, J.E.; Polya, G. (1952년), 불평등, Cambridge Matheical Library (2. Ed.), Cambridge: 케임브리지 대학 출판부, ISBN 0-521-05206-8, MR0046395, Zbl 0047.05302, 섹션 2.18, 정리 45.