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:
  
<math>\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}}</math>
+
<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]].
 
[[Image:RMS-AM-GM-HM.gif|frame|right|The inequality is clearly shown in this diagram for <math>n=2</math>]]
 
  
 
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  
<math>(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</math> 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.
+
<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]]; <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]];
 +
<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:
<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>;
+
<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>
<math>F\left(\frac{x_1+\cdots+x_n}{n}\right)\le \frac {F(x_1)+\cdots+F(x_n)}{n}</math>
+
<cmath>\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}</cmath>
 
 
 
 
<math>\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}</math>
 
 
 
 
 
 
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>
<math>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n} \blacksquare.</math>
 
  
  
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Therefore the original inequality is true.
 
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.
+
===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 $x_1,\ldots,x_n$ 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.

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.

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

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 $n=2$

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