Difference between revisions of "Maclaurin's Inequality"

 
 
(3 intermediate revisions by 2 users not shown)
Line 9: Line 9:
 
</math>,
 
</math>,
 
</center>
 
</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 ==
 
== 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>
 
<center>
 
<math>
 
<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]].
 
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]]
 
* [[Inequality]]
 
* [[Newton's Inequality]]
 
* [[Newton's Inequality]]
 
* [[Symmetric sum]]
 
* [[Symmetric sum]]
 +
 +
[[Category:Algebra]]
 +
[[Category:Inequalities]]

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.

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.

See Also

Invalid username
Login to AoPS