Difference between revisions of "AM-GM Inequality"

Revision as of 15:20, 30 November 2021

In Algebra, the AM-GM Inequality, or more formally the Inequality of Arithmetic and Geometric Means, states that the arithmetic mean is greater than or equal to the geometric mean of any list of nonnegative reals; furthermore, equality holds if and only if every real in the list is the same.

In symbols, the inequality states that for any real numbers $x_1, x_2, \ldots, x_n \geq 0$ , $\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}$ with equality if and only if $x_1 = x_2 = \cdots = x_n$ .

NOTE: This article is a work-in-progress and meant to replace the Arithmetic mean-geometric mean inequality article, which is of poor quality.

Proofs

Main article: Proofs of AM-GM

All known proofs of AM-GM use either induction or other inequalities. Its proof is far more complicated than its usage in introductory competitions, which makes learning it not recommended for students new to proofs. Listed here is proof of AM-GM that utilizes Cauchy Induction

Base Case

The smallest nontrivial case of AM-GM is in two variables. Because it is a square (or by the Trivial Inequality), $(x-y)^2 \geq 0,$ with equality if and only if $x-y=0$ , or $x=y$ . Then because $x$ and $y$ are nonnegative, we can perform the following manipulations: $x^2 - 2xy + y^2 \geq 0$ $x^2 + 2xy + y^2 \geq 4xy$ $\frac{(x+y)^2}{4} \geq xy$ $\frac{x+y}{2} \geq \sqrt{xy}.$ This completes the proof of the base case.

Powers of Two

Backwards Step

Generalizations

The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality is another generalization of AM-GM.

Weighted AM-GM Inequality

The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights $\omega_1, \omega_2, \ldots, \omega_n \geq 0$ such that $\omega_1 + \omega_2 + \cdots + \omega_n = \omega$ , $\frac{\omega_1 x_1 + \omega_2 x_2 + \cdots + \omega_n x_n}{\omega} \geq \sqrt[\omega]{x_1^{\omega_1} x_2^{\omega_2} \cdots x_n^{\omega_n}},$ with equality if and only if $x_1 = x_2 = \cdots = x_n$ . When $\omega_1 = \omega_2 = \cdots = \omega_n = 1/n$ , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.

Mean Inequality Chain

Main article: Mean Inequality Chain

The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that $\sqrt{\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}} \geq \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} \geq \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}},$ with equality if and only if $x_1 = x_2 = \cdots = x_n$ . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.

Power Mean Inequality

Main article: Power Mean Inequality

The Power Mean Inequality relates every power mean of a list of nonnegative reals. The power mean $M(p)$ is defined as follows: $M(p) = \begin{cases} (\frac{x_1^p + x_2^p + \cdots + x_n^p}{n})^\frac{1}{p} &\text{if } p \neq 0 \\ \sqrt[n]{x_1 x_2 \cdots x_n} &\text{if } p = 0. \end{cases}$ The Power Mean inequality then states that if $a>b$ , then $M(a) \geq M(b)$ , with equality holding if and only if $x_1 = x_2 = \cdots = x_n.$ Plugging $p=1, 0$ into this inequality reduces it to AM-GM, and $p=2, 1, 0, -1$ gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.

Introductory examples

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Intermediate examples

WIP

Olympiad examples

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@@ Line 10: / Line 10: @@
 === Base Case ===
-The smallest nontrivial case of AM-GM is in two variables. By the [[Trivial Inequality]], <cmath>(x-y)^2 \geq 0,</cmath> with equality if and only if <math>x-y=0</math>, or <math>x=y</math>. Then because <math>x</math> and <math>y</math> are nonnegative, we can perform the following manipulations: <cmath>x^2 - 2xy + y^2 \geq 0</cmath> <cmath>x^2 + 2xy + y^2 \geq 4xy</cmath> <cmath>\frac{(x+y)^2}{4} \geq xy</cmath> <cmath>\frac{x+y}{2} \geq \sqrt{xy}.</cmath> This completes the proof of the base case.
+The smallest nontrivial case of AM-GM is in two variables. Because it is a square (or by the [[Trivial Inequality]]), <math>(x-y)^2 \geq 0,</math> with equality if and only if <math>x-y=0</math>, or <math>x=y</math>. Then because <math>x</math> and <math>y</math> are nonnegative, we can perform the following manipulations: <cmath>x^2 - 2xy + y^2 \geq 0</cmath> <cmath>x^2 + 2xy + y^2 \geq 4xy</cmath> <cmath>\frac{(x+y)^2}{4} \geq xy</cmath> <cmath>\frac{x+y}{2} \geq \sqrt{xy}.</cmath> This completes the proof of the base case.
 === Powers of Two ===

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