Difference between revisions of "AM-GM Inequality"

Latest revision as of 14:32, 22 February 2024

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 $x_1 = x_2 = \cdots = x_n$ .

The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.

Proofs

Main article: Proofs of AM-GM

All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for $2$ , uses induction to extend this to all powers of $2$ , and then shows that assuming the result for $n$ implies it holds for $n-1$ .

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 1: / Line 1: @@
-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 [[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 <math>x_1,  x_2, \ldots, x_n \geq 0</math>, <cmath>\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}</cmath> with equality if and only if <math>x_1 = x_2 = \cdots = x_n</math>.
+In symbols, the inequality states that for any real numbers <math>x_1,  x_2, \ldots, x_n \geq 0</math>, <cmath>\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}</cmath> with [[equality condition | equality]] if and only if <math>x_1 = x_2 = \cdots = x_n</math>.
-'''NOTE''': This article is a work-in-progress and meant to replace the [[Arithmetic mean-geometric mean inequality]] article, which is of poor quality.
+The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
 == Proofs ==
 {{Main|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 <math>n</math> implies <math>n-1</math>.
+All known proofs of AM-GM use [[induction]] or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes [[Cauchy Induction]], a variant of induction where one proves a result for <math>2</math>, uses induction to extend this to all powers of <math>2</math>, and then shows that assuming the result for <math>n</math> implies it holds for <math>n-1</math>.
 == Generalizations ==
@@ 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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