Maclaurin's Inequality: Difference between revisions

Latest revision as of 15:48, 29 December 2021

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

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.

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.

@@ Line 9: / Line 9: @@
 </math>,
 </center>
-with equality exactly when all the <math> \displaystyle x_i </math> are equal.
+with equality exactly when all the <math>x_i </math> are equal.
 == Proof ==
-By the lemma from [[Newton's Inequality]], it suffices to show that for any <math> \displaystyle n </math>,
+By the lemma from [[Newton's Inequality]], it suffices to show that for any <math>n </math>,
 <center>
 <math>
@@ Line 27: / Line 27: @@
 Since the [[geometric mean]] of <math> 1/x_1, \ldots, 1/x_n </math> is 1, the inequality is true by [[AM-GM]].
-== Resources ==
+== See Also ==
 * [[Inequality]]
 * [[Newton's Inequality]]
 * [[Symmetric sum]]
+[[Category:Algebra]]
+[[Category:Inequalities]]