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Maclaurin's Inequality

From AoPSWiki

Maclaurin's Inequality is an inequality in symmetric polynomials. For notation and background, we refer to Newton's Inequality.

Statement

For non-negative $x_1, \ldots, x_n$ ,

$d_1 \ge d_2^{1/2} \ge \ldots \ge d_n^{1/n}$ ,

with equality exactly when all the $x_i$ are equal.

Proof

By the lemma from Newton's Inequality, it suffices to show that for any $n$ ,

$d_{n-1}^{1/(n-1)} \ge d_{n}^{1/n}$ .

Since this is a homogenous inequality, we may normalize so that $d_n = \prod x_i = 1$ . We then transform the inequality to

$\frac{\sum 1/x_i}{n} \ge 1^{\frac{n-1}{n}} = 1$ .

Since the geometric mean of $1/x_1, \ldots, 1/x_n$ is 1, the inequality is true by AM-GM.

Resources

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/Maclaurin%27s_Inequality"

Categories: Inequality | Theorems

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

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