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| − | The '''Arithmetic Mean-Geometric Mean Inequality''' ('''AM-GM''' or '''AMGM''') is an elementary [[inequality]], and is generally one of the first ones taught in inequality courses.
| + | #REDIRECT[[AM-GM Inequality]] |
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| − | == Theorem ==
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| − | AM-GM states that for any [[set]] of [[nonnegative]] [[real number]]s, the [[arithmetic mean]] of the set is greater than or [[equal]] to the [[geometric mean]] of the set. Algebraically, this is expressed as follows.
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| − | For a set of nonnegative real numbers <math>a_1,a_2,\ldots,a_n</math>, the following always holds:
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| − | <cmath> \frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n} </cmath>
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| − | Using the shorthand notation for [[summation]]s and [[product]]s:
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| − | <cmath> \sum_{i=1}^{n}\frac{a_i}{n} \geq \prod\limits_{i=1}^{n}a_i^{\frac{1}{n}} . </cmath>
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| − | For example, for the set <math>\{9,12,54\}</math>, the arithmetic mean, 25, is greater than the geometric mean, 18; AM-GM guarantees this is always the case.
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| − | The [[equality condition]] of this [[inequality]] states that the arithmetic mean and geometric mean are equal [[iff|if and only if]] all members of the set are equal.
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| − | AM-GM can be used fairly frequently to solve [[Olympiad]]-level inequality problems, such as those on the [[United States of America Mathematics Olympiad | USAMO]] and [[International Mathematics Olympiad | IMO]].
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| − | === Proof ===
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| − | See here: [[Proofs of AM-GM]].
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| − | === Weighted Form ===
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| − | The weighted form of AM-GM is given by using [[weighted average]]s. For example, the weighted arithmetic mean of <math>x</math> and <math>y</math> with <math>3:1</math> is <math>\frac{3x+1y}{3+1}</math> and the geometric is <math>\sqrt[3+1]{x^3y}</math>.
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| − | AM-GM applies to weighted averages. Specifically, the '''weighted AM-GM Inequality''' states that if <math>a_1, a_2, \dotsc, a_n</math> are nonnegative real numbers, and <math>\lambda_1, \lambda_2, \dotsc, \lambda_n</math> are nonnegative real numbers (the "weights") which sum to 1, then
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| − | <cmath> \lambda_1 a_1 + \lambda_2 a_2 + \dotsb + \lambda_n a_n \ge a_1^{\lambda_1} a_2^{\lambda_2} \dotsm a_n^{\lambda_n}, </cmath>
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| − | or, in more compact notation,
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| − | <cmath> \sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i} . </cmath>
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| − | Equality holds if and only if <math>a_i = a_j</math> for all integers <math>i, j</math> such that <math>\lambda_i \neq 0</math> and <math>\lambda_j \neq 0</math>.
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| − | We obtain the unweighted form of AM-GM by setting <math>\lambda_1 = \lambda_2 = \dotsb = \lambda_n = 1/n</math>.
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| − | ==Extensions==
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| − | * The [[power mean inequality]] is a generalization of AM-GM which places the arithemetic and geometric means on a continuum of different means.
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| − | * The [[root-square-mean arithmetic-mean geometric-mean harmonic-mean inequality]] is special case of the power mean inequality.
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| − | * Kedlaya also extended it greatly doing some stuff like the arithmetic mean of the sequence of geometric means is at least the geometric mean of the sequence of arithmetic means or something like that.
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| − | ==Problems==
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| − | === Introductory ===
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| − | * Demonstrate that if <math>a_1a_2\cdots a_n=1</math> then <math>a_1+a_2+\cdots +a_n\ge n</math>.
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| − | === Intermediate ===
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| − | * Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.
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| − | ([[1983 AIME Problems/Problem 9|Source]])
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| − | === Olympiad ===
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| − | * Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers. Prove that
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| − | <cmath> (a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3 . </cmath>
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| − | ([[2004 USAMO Problems/Problem 5|Source]])
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| − | == See Also ==
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| − | * [[RMS-AM-GM-HM]]
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| − | * [[Algebra]]
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| − | * [[Inequalities]]
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| − | [[Category:Inequality]]
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| − | [[Category:Theorems]]
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