Difference between revisions of "Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality"

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Suppose we let <math>F(x)=x^2</math> (We know that <math>F(x)</math> is convex because <math>F'(x)=2x</math> and therefore <math>F''(x)=2>0</math>).  
 
Suppose we let <math>F(x)=x^2</math> (We know that <math>F(x)</math> is convex because <math>F'(x)=2x</math> and therefore <math>F''(x)=2>0</math>).  
 
We have:
 
We have:
<math>F\left(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\right)\le \frac1)}{n}+\cdots+\frac{F(x_n)}{n}</math>;
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<math>F\left(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\right)\le \frac{F(x_1)}{n}+\cdots+\frac{F(x_n)}{n}</math>;
  
 
Factoring out the <math>\frac{1}{n}</math> yields:
 
Factoring out the <math>\frac{1}{n}</math> yields:

Revision as of 13:44, 24 December 2014

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 $\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}$.

Alternatively, the RMS-AM can be proved using Jensen's inequality: Suppose we let $F(x)=x^2$ (We know that $F(x)$ is convex because $F'(x)=2x$ and therefore $F''(x)=2>0$). We have: $F\left(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\right)\le \frac{F(x_1)}{n}+\cdots+\frac{F(x_n)}{n}$;

Factoring out the $\frac{1}{n}$ yields:

$F\left(\frac{x_1+\cdots+x_n}{n}\right)\le \frac {F(x_1)+\cdots+F(x_n)}{n}$


$\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}$


Taking the square root to both sides (remember that both are positive):

$\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n} \blacksquare.$


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.

The Root Mean Square is also known as the quadratic mean, and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality.


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