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

The '''Root-Mean Power-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality''' (RMP-AM-GM-HM) or '''Exponential Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality''' (EM-AM-GM-HM), is an [[inequality]] of the [[root-mean power]], [[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[n_1]{\frac{x_1^{n_1}+\cdots+x_a^{n_1}}{a}} \ge \frac{x_1+\cdots+x_n}{n} \ge \sqrt[n_2]{\frac{x_1^{n_2}+\cdots+x_a^{n_2}}{a}} \ge \sqrt[n]{x_1\cdots x_n} \ge \sqrt[n_3]{\frac{x_1^{n_3}+\cdots+x_a^{n_3}}{a}} \ge \frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}} \ge \sqrt[n_4]{\frac{x_1^{n_4}+\cdots+x_a^{n_4}}{a}}</cmath> where <math>n_1>1,~~0<n_2<1,~~-1<n_3<0,~~n_4<-1</math>, and <math>n</math> is the root mean power.

The geometric mean is the theoretical existence if the root mean power equals 0, which we couldn't calculate using radicals because the 0th root of any number is undefined when the number's absolute value is greater than or equal to 1.

The quadratic mean's root mean power is 2, the arithmetic mean's root mean power is 1, as <math>\frac{x_1+\cdots+x_n}{n}=\sqrt[1]{\frac{x_1^1+\cdots+x_a^1}{a}}</math> and the harmonic mean's root mean power is -1 as <math>\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}=\sqrt[-1]{\frac{x_1^{-1}+\cdots+x_a^{-1}}{a}}</math>. Similarly, their is a root mean cube, or cubic mean, whose root mean power equals 3 which is <math>\sqrt[3]{\frac{x_1^3+\cdots+x_a^3}{a}}</math>

When the root mean power approaches <math>\infty</math>, the mean approaches the highest number. When the root mean power reaches <math>-\infty</math>, the mean approaches the lowest number.

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]], and is also known as the Mean Inequality Chain.

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 Power of 2 is also known as the [[quadratic mean]], and the inequality is therefore sometimes known as the EM-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 Proofs===

<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>

[https://www.desmos.com/calculator/3lveujduax Desmos Sliders]

(Note how the RMS is "sandwiched" between the minimum and the maximum)

[[Category:Algebra]]
[[Category:Inequalities]]