title: Maclaurin's inequality
text: # Maclaurin's inequality
  
  In mathematics, Maclaurin's inequality, named after Colin Maclaurin, is a refinement of the inequality of arithmetic and geometric means.
  
  Let $a_1,$ $a_2,$ ..., $a_n$ 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 $a_1,$ $a_2,$ ..., $a_n,$ that is, the sum of all products of k of the numbers $a_1,$ $a_2,$ ..., $a_n$ 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:
  
      ${\begin{aligned}&\quad {\frac {a_{1}+a_{2}+a_{3}+a_{4}}{4}}\\[8pt]&\geq {\sqrt {\frac {a_{1}a_{2}+a_{1}a_{3}+a_{1}a_{4}+a_{2}a_{3}+a_{2}a_{4}+a_{3}a_{4}}{6}}}\\[8pt]&\geq {\sqrt[{3}]{\frac {a_{1}a_{2}a_{3}+a_{1}a_{2}a_{4}+a_{1}a_{3}a_{4}+a_{2}a_{3}a_{4}}{4}}}\\[8pt]&\geq {\sqrt[{4}]{a_{1}a_{2}a_{3}a_{4}}}.\end{aligned}}$
  
  Maclaurin's inequality can be proved using Newton's inequalities or generalised Bernoulli's inequality.
  
  