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Difference between revisions of "Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality"
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== Proof ==
 
== Proof ==
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The inequality <math>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}</math> is a direct consequence of the [[Cauchy-Schwarz Inequality]]; <math>(x_1^2+x_2^2+\cdots +x_n^2)(1+1+\cdots +1)\geq (x_1+x_2+\cdots +x_n)^2</math>, so <math>\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}\geq \left(\frac{x_1+x_2+\cdots +x_n}{n}\right)^2</math>, so <math>sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}</math>.
+
The inequality <math>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}</math> is a direct consequence of the [[Cauchy-Schwarz Inequality]]; <math>(x_1^2+x_2^2+\cdots +x_n^2)(1+1+\cdots +1)\geq (x_1+x_2+\cdots +x_n)^2</math>, so <math>\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}\geq \left(\frac{x_1+x_2+\cdots +x_n}{n}\right)^2</math>, so <math>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}</math>.
  
 
The inequality <math>\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}</math> is called the AM-GM inequality, and proofs can be found [[Proofs of AM-GM|here]].
 
The inequality <math>\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}</math> is called the AM-GM inequality, and proofs can be found [[Proofs of AM-GM|here]].
  
−
The inequality <math>\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}</math> is a direct consequence of AM-GM; <math>\frac{\sum_{i=1}^{n}\sqrt[n]{\frac{x_1x_2\cdots x_n}{x_i^n}}}{n}\geq 1</math>, so <math>\sqrt[n]{x_1x_2\cdots x_n}\frac{\sum_{i=1}^{n}\frac{1}{x_i}}}{n}\geq 1</math>, so <math>\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}</math>.
+
The inequality <math>\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}</math> is a direct consequence of AM-GM; <math>\frac{\sum_{i=1}^{n}\sqrt[n]{\frac{x_1x_2\cdots x_n}{x_i^n}}}{n}\geq 1</math>, so <math>\sqrt[n]{x_1x_2\cdots x_n}\frac{\sum_{i=1}^{n}\frac{1}{x_i}}{n}\geq 1</math>, so <math>\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}</math>.
  
 
Therefore the original inequality is true.
 
Therefore the original inequality is true.

Revision as of 13:40, 21 July 2010

The Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality (RMS-AM-GM-HM), is an inequality of the root-mean square, arithmetic mean, geometric mean, and harmonic mean of a set of positive real numbers $x_1,\ldots,x_n$ that says:

$\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}$

with equality if and only if $x_1=x_2=\cdots=x_n$. This inequality can be expanded to the power mean inequality.

The inequality is clearly shown in this diagram for $n=2$

As a consequence we can have the following inequality: If $x_1,x_2,\cdots,x_n$ are positive reals, then $(x_1+x_2+\cdots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots \frac{1}{x_n}\right) \geq n^2$ with equality if and only if $x_1=x_2=\cdots=x_n$; which follows directly by cross multiplication from the AM-HM inequality.This is extremely useful in problem solving.

Proof

The inequality $\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}$ is a direct consequence of the Cauchy-Schwarz Inequality; $(x_1^2+x_2^2+\cdots +x_n^2)(1+1+\cdots +1)\geq (x_1+x_2+\cdots +x_n)^2$, so $\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}\geq \left(\frac{x_1+x_2+\cdots +x_n}{n}\right)^2$, so $\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}$.

The inequality $\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}$ is called the AM-GM inequality, and proofs can be found here.

The inequality $\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}$ is a direct consequence of AM-GM; $\frac{\sum_{i=1}^{n}\sqrt[n]{\frac{x_1x_2\cdots x_n}{x_i^n}}}{n}\geq 1$, so $\sqrt[n]{x_1x_2\cdots x_n}\frac{\sum_{i=1}^{n}\frac{1}{x_i}}{n}\geq 1$, so $\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}$.

Therefore the original inequality is true.

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