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:

$S_k = \frac{\displaystyle \sum_{ 1\leq i_1 < \cdots < i_k \leq n}a_{i_1} a_{i_2} \cdots a_{i_k}}{\displaystyle {n \choose k}}.$

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 $\scriptstyle {n\choose k}.$

Maclaurin's inequality is the following chain of inequalities:

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

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:

$\begin{align} & {} \quad \frac{a_1+a_2+a_3+a_4}{4} \\[8pt] & {} \ge \sqrt{\frac{a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4}{6}} \\[8pt] & {} \ge \sqrt[3]{\frac{a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4}{4}} \\[8pt] & {} \ge \sqrt[4]{a_1a_2a_3a_4}. \end{align}$

Maclaurin's inequality can be proved using the Newton's inequalities.

See also

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

References

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

This article incorporates material from MacLaurin's Inequality on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.