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| − | 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:
| + | #REDIRECT [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality]] |
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| − | <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>
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| − | 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.
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| − | As a consequence we can have the following inequality:
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| − | If <math>x_1,x_2,\cdots,x_n</math> are positive reals, then
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| − | <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>
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| − | 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.
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| − | 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.
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| − | == Proof ==
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| − | 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]];
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| − | <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>
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| − | <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>
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| − | <cmath>\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}</cmath>
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| − | Alternatively, the RMS-AM can be proved using Jensen's inequality:
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| − | 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>).
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| − | We have:
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| − | <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>
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| − | Factoring out the <math>\frac{1}{n}</math> yields:
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| − | <cmath>F\left(\frac{x_1+\cdots+x_n}{n}\right)\le \frac {F(x_1)+\cdots+F(x_n)}{n}</cmath>
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| − | <cmath>\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}</cmath>
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| − | Taking the square root to both sides (remember that both are positive):
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| − | <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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| − | 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]].
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| − | 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>.
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| − | Therefore the original inequality is true.
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| − | ===Geometric Proof===
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| − | <asy>size(250);
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| − | 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);
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| − | draw(B--O--C--arc(O,C,A)--O--R--P); rightanglemark(O,P,R);
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| − | draw(O--B,red);
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| − | draw(P--Q,blue);
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| − | draw(B--P,green);
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| − | draw(R--Q,magenta);
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| − | draw((A-(0,0.05))--(P-(0,0.05)),Arrows);
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| − | draw((P-(0,0.05))--(C-(0,0.05)),Arrows);
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| − | label("AM",(O+B)/2,W,red);
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| − | label("GM",(P+Q)/2,E,blue);
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| − | label("HM",(R+Q)/2,unit(P-R),magenta);
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| − | label("RMS",(3B+P)/4,unit(foot(O,B,P)),green);
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| − | label("$a$",(A+P)/2,3*S);
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| − | label("$b$",(P+C)/2,3*S);</asy>
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| − | The inequality is clearly shown in this diagram for <math>n=2</math>
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| − | {{stub}}
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| − | [[Category:Inequality]]
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| − | [[Category:Theorems]]
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