View source for Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality

The '''Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality''' (RMS-AM-GM-HM) or '''Quadratic Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality''' (QM-AM-GM-HM), is an [[inequality]] of the [[root-mean square]], [[arithmetic mean]], [[geometric mean]], and [[harmonic mean]] of a set of  [[positive]] [[real number]]s <math>x_1,\ldots,x_n</math> that says:

<cmath>\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}}</cmath>

with equality if and only if <math>x_1=x_2=\cdots=x_n</math>.  This inequality can be expanded to the [[power mean inequality]].

As a consequence we can have the following inequality:
If <math>x_1,x_2,\cdots,x_n</math> are positive reals, then 
<cmath>(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</cmath>
with equality if and only if <math>x_1=x_2=\cdots=x_n</math>; which follows directly by cross multiplication from the AM-HM inequality. This is extremely useful in problem solving.

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. 

== Proof ==

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]];
<cmath>(x_1^2+x_2^2+\cdots +x_n^2)(1+1+\cdots +1)\geq (x_1+x_2+\cdots +x_n)^2</cmath>
<cmath>\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</cmath>
<cmath>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}</cmath>
Alternatively, the RMS-AM can be proved using Jensen's inequality:
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:
<cmath>F\left(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\right)\le \frac{F(x_1)}{n}+\cdots+\frac{F(x_n)}{n}</cmath>
Factoring out the <math>\frac{1}{n}</math> yields:
<cmath>F\left(\frac{x_1+\cdots+x_n}{n}\right)\le \frac {F(x_1)+\cdots+F(x_n)}{n}</cmath>
<cmath>\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}</cmath>
Taking the square root to both sides (remember that both are positive):
<cmath>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n} \blacksquare.</cmath>



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>.

Therefore the original inequality is true.

===Geometric Proof===

<asy>size(250);
pair O=(0,0),A=(-1,0),B=(0,1),C=(1,0),P=(1/2,0),Q=(1/2,sqrt(3)/2),R=foot(P,Q,O);
draw(B--O--C--arc(O,C,A)--O--R--P); rightanglemark(O,P,R);
draw(O--B,red);
draw(P--Q,blue);
draw(B--P,green);
draw(R--Q,magenta);
draw((A-(0,0.05))--(P-(0,0.05)),Arrows);
draw((P-(0,0.05))--(C-(0,0.05)),Arrows);
label("AM",(O+B)/2,W,red);
label("GM",(P+Q)/2,E,blue);
label("HM",(R+Q)/2,unit(P-R),magenta);
label("RMS",(3B+P)/4,unit(foot(O,B,P)),green);
label("$a$",(A+P)/2,3*S);
label("$b$",(P+C)/2,3*S);</asy>

The inequality is clearly shown in this diagram for <math>n=2</math>

{{stub}}
[[Category:Inequality]]
[[Category:Theorems]]