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Muirhead's inequality: Difference between revisions

Revision as of 10:11, 8 February 2024

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_nby

${\displaystyle [$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 meanofx₁, ..., x_n.
• For a = (1/n, ..., 1/n), the a-mean is the geometric meanofx₁, ..., 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

Ann × n matrix Pisdoubly 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

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

by writing it using the symmetric-sum notation as

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

which is the same as

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

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

See also

• Inequality of arithmetic and geometric means
• Doubly stochastic matrix
• Monomial symmetric polynomial

Notes

References

• Combinatorial Theory by John N. Guidi, based on lectures given by Gian-Carlo Rota in 1998, MIT Copy Technology Center, 2002.
• Kiran Kedlaya, A < B (A less than B), a guide to solving inequalities
• Muirhead's theorematPlanetMath.
• Hardy, G.H.; Littlewood, J.E.; Pólya, G. (1952), Inequalities, Cambridge Mathematical Library (2. ed.), Cambridge: Cambridge University Press, ISBN 0-521-05206-8, MR0046395, Zbl 0047.05302, Section 2.18, Theorem 45.