Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mean Inequality Chain"
Line 59: Line 59:
  
 
{{stub}}
 
{{stub}}
−
[[Category:Inequality]]
+
[[Category:Algebra]]
−
[[Category:Theorems]]
+
[[Category:Inequalities]]

Revision as of 15:49, 29 December 2021

note -- taken from https://artofproblemsolving.com/wiki/index.php/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 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, and is also known as the Mean Inequality Chain.

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 $n=2$

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mean_Inequality_Chain&oldid=168868"