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=(a1,...,an)
define the "a-mean" [''a''] of positive real numbers x1, ..., xn by
[a]= | 1 |
n! |
\sum\sigma
a1 | |
x | |
\sigma1 |
…
an | |
x | |
\sigman |
,
where the sum extends over all permutations σ of .
ma(x1,...,xn)
[a]=
k1! … kl! | |
n! |
ma(x1,...,xn),
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
a1+ … +an=1
1/(a1+ … +an) | |
[a] |
See main article: 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.
Muirhead's inequality states that [''a''] ≤ [''b''] for all x such that xi > 0 for every i ∈ 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).
Because of the symmetry of the sum, no generality is lost by sorting the exponents into decreasing order:
a1\geqa2\geq … \geqan
b1\geqb2\geq … \geqbn.
Then the existence of a doubly stochastic matrix P such that a = Pb is equivalent to the following system of inequalities:
\begin{align} a1&\leqb1\\ a1+a2&\leqb1+b2\\ a1+a2+a3&\leqb1+b2+b3\\ &\vdots\\ a1+ … +an-1&\leqb1+ … +bn-1\\ a1+ … +an&=b1+ … +bn. \end{align}
(The last one is an equality; the others are weak inequalities.)
The sequence
b1,\ldots,bn
a1,\ldots,an
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 (
\alpha1,\ldots,\alphan
\sumsym
\alpha1 | |
x | |
1 |
…
\alphan | |
x | |
n |
This notation requires developing every permutation, developing an expression made of n! monomials, for instance:
\begin{align} \sumsymx3y2z0&=x3y2z0+x3z2y0+y3x2z0+y3z2x0+z3x2y0+z3y2x0\\ &=x3y2+x3z2+y3x2+y3z2+z3x2+z3y2 \end{align}
See main article: Inequality of arithmetic and geometric means.
Let
aG=\left(
1 | |
n |
,\ldots,
1 | |
n |
\right)
aA=(1,0,0,\ldots,0).
We have
\begin{align} aA1=1&>aG1=
1 | |
n, |
\\ aA1+aA2=1&>aG1+aG2=
2 | |
n, |
\\ &\vdots\\ aA1+ … +aAn&=aG1+ … +aGn=1. \end{align}
[''a<sub>A</sub>''] ≥ [''a<sub>G</sub>''],which is
1 | |
n! |
1 | |
(x | |
1 |
⋅
0 | |
x | |
2 |
…
0 | |
x | |
n |
+ … +
0 | |
x | |
1 |
…
1) | |
x | |
n |
(n-1)!\geq
1 | |
n! |
(x1 ⋅ … ⋅
1/n | |
x | |
n) |
n!
We seek to prove that x2 + y2 ≥ 2xy by using bunching (Muirhead's inequality).We transform it in the symmetric-sum notation:
\sumsymx2y0\ge\sumsymx1y1.
The sequence (2, 0) majorizes the sequence (1, 1), thus the inequality holds by bunching.
Similarly, we can prove the inequality
x3+y3+z3\ge3xyz
by writing it using the symmetric-sum notation as
\sumsymx3y0z0\ge\sumsymx1y1z1,
which is the same as
2x3+2y3+2z3\ge6xyz.
Since the sequence (3, 0, 0) majorizes the sequence (1, 1, 1), the inequality holds by bunching.