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

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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 number]]s <math>x_1,\ldots,x_n</math> that says: | 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 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]]. | 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: | As a consequence we can have the following inequality: | ||

If <math>x_1,x_2,\cdots,x_n</math> are positive reals, then | 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 == | == 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: | 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>). | 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: | ||

− | < | + | <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: | 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): | 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> | |

− | < | ||

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Therefore the original inequality is true. | 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}} | {{stub}} | ||

[[Category:Inequality]] | [[Category:Inequality]] | ||

[[Category:Theorems]] | [[Category:Theorems]] |

## Latest revision as of 07:35, 28 July 2019

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 that says:

with equality if and only if . This inequality can be expanded to the power mean inequality.

As a consequence we can have the following inequality: If are positive reals, then with equality if and only if ; 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 is a direct consequence of the Cauchy-Schwarz Inequality; Alternatively, the RMS-AM can be proved using Jensen's inequality: Suppose we let (We know that is convex because and therefore ). We have: Factoring out the yields: Taking the square root to both sides (remember that both are positive):

The inequality is called the AM-GM inequality, and proofs can be found here.

The inequality is a direct consequence of AM-GM; , so , so .

Therefore the original inequality is true.

### Geometric Proof

The inequality is clearly shown in this diagram for

*This article is a stub. Help us out by expanding it.*