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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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| − | ==Problems==
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| − | === Introductory ===
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| − | * 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>.
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| − | * Find the maximum of <math>2 - a - \frac{1}{2a}</math> for all positive <math>a</math>. [[Solution to AM - GM Introductory Problem 2]]
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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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