Maclaurin's inequality

In mathematics, Maclaurin's inequality, named after Colin Maclaurin, is a refinement of the inequality of arithmetic and geometric means. Let a₁, a₂, ..., a_n be positive real numbers, and for k = 1, 2, ..., n define the averages S_k as follows:

$S_k=\frac{\sum _{1\le i_1<\cdots <i_k\le n}{a}_{i_1}{a}_{i_2}\cdots a_{i_k}}{(\genfrac{}{}{0ex}{}{n}{k})}.$

The numerator of this fraction is the elementary symmetric polynomial of degree k in the n variables a₁, a₂, ..., a_n, that is, the sum of all products of k of the numbers a₁, a₂, ..., a_n with the indices in increasing order. The denominator is the number of terms in the numerator, the binomial coefficient $(\genfrac{}{}{0ex}{}{n}{k})$$.$

Maclaurin's inequality is the following chain of inequalities:

$S_1\ge \sqrt{S_2}\ge \sqrt[3]{S_3}\ge \cdots \ge \sqrt[n]{S_n}$

with equality if and only if all the a_i are equal.

For n = 2, this gives the usual inequality of arithmetic and geometric means of two numbers. Maclaurin's inequality is well illustrated by the case n = 4:

$\begin{array}{rl}& \frac{a_1+a_2+a_3+a_4}{4}\\ & \ge \sqrt{\frac{a_1{a}_2+a_1{a}_3+a_1{a}_4+a_2{a}_3+a_2{a}_4+a_3{a}_4}{6}}\\ & \ge \sqrt[3]{\frac{a_1{a}_2{a}_3+a_1{a}_2{a}_4+a_1{a}_3{a}_4+a_2{a}_3{a}_4}{4}}\\ & \ge \sqrt[4]{a_1{a}_2{a}_3{a}_4}.\end{array}$

Maclaurin's inequality can be proved using Newton's inequalities or generalised Bernoulli's inequality.

See also

• Newton's inequalities
• Muirhead's inequality
• Generalized mean inequality
• Bernoulli's inequality

References

• Biler, Piotr; Witkowski, Alfred (1990). Problems in mathematical analysis. New York, N.Y.: M. Dekker. ISBN 0-8247-8312-3.