Difference between revisions of "AM-GM Inequality"

Revision as of 15:50, 29 December 2021

In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean; furthermore, the two means are equal if and only if every number 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, more advanced inequalities. Its proof is far more complicated than its usage in introductory competitions; consequentially, learning it is not recommended to students new to proofs. The most elementary proof of AM-GM utilizes Cauchy Induction, a variant of induction that involves proving a result for two, then using induction to prove it for all powers of two, and then a backward step where $n$ implies $n-1$ .

Generalizations

The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations 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 all the different power means of a list of nonnegative reals. The power mean $M(p)$ is defined as follows: $M(p) = \begin{cases} \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n}\right)^\frac{1}{p} &\text{if } p \neq 0 \\ \sqrt[n]{x_1 x_2 \cdots x_n} &\text{if } p = 0. \end{cases}$ 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.

Problems

Introductory

For nonnegative real numbers $a_1,a_2,\cdots a_n$ , demonstrate that if $a_1a_2\cdots a_n=1$ then $a_1+a_2+\cdots +a_n\ge n$ . (Solution)
Find the maximum of $2 - a - \frac{1}{2a}$ for all positive $a$ . (Solution)

Intermediate

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$ .

(Source)

Olympiad

Let $a$ , $b$ , and $c$ be positive real numbers. Prove that

$(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3 .$ (Source)

@@ Line 23: / Line 23: @@
 The '''Power Mean Inequality''' relates all the different power means of a list of nonnegative reals. The power mean <math>M(p)</math> is defined as follows: <cmath>M(p) = \begin{cases} \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n}\right)^\frac{1}{p} &\text{if } p \neq 0 \\ \sqrt[n]{x_1 x_2 \cdots x_n} &\text{if } p = 0. \end{cases}</cmath> The Power Mean inequality then states that if <math>a>b</math>, then <math>M(a) \geq M(b)</math>, with equality holding if and only if <math>x_1 = x_2 = \cdots = x_n.</math> Plugging <math>p=1, 0</math> into this inequality reduces it to AM-GM, and <math>p=2, 1, 0, -1</math> gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
+== Problems ==
-== Introductory examples ==
+=== Introductory ===
-WIP
+* For nonnegative real numbers <math>a_1,a_2,\cdots a_n</math>, demonstrate that if <math>a_1a_2\cdots a_n=1</math> then <math>a_1+a_2+\cdots +a_n\ge n</math>. ([[Solution to AM - GM Introductory Problem 1|Solution]])
+* Find the maximum of <math>2 - a - \frac{1}{2a}</math> for all positive <math>a</math>. ([[Solution to AM - GM Introductory Problem 2|Solution]])
-== Intermediate examples ==
+=== Intermediate ===
-WIP
+* Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.
+([[1983 AIME Problems/Problem 9|Source]])
-== Olympiad examples ==
+=== Olympiad ===
-WIP
+* Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers.  Prove that
+<cmath> (a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3 . </cmath>
-== More Problems ==
+([[2004 USAMO Problems/Problem 5|Source]])
-WIP
 == See Also ==
+* [[Proofs of AM-GM]]
 * [[Mean Inequality Chain]]
 * [[Power Mean Inequality]]
 * [[Cauchy-Schwarz Inequality]]
 * [[Inequality]]
+[[Category:Algebra]]
+[[Category:Inequalities]]
+[[Category:Definition]]

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