Arithmetic Mean-Geometric Mean Inequality
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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.
Contents
Theorem
The AM-GM states that for any set of nonnegative real numbers, the arithmetic mean of the set is greater than or equal to the geometric mean of the set. Algebraically, this is expressed as follows.
For a set of nonnegative real numbers , the following always holds: Using the shorthand notation for summations and products:
\[\sum_{i=1}^{n}a_i}/n \geq \prod\limits_{i=1}^{n}a_i^{1/n} .\] (Error compiling LaTeX. ! Extra }, or forgotten $.)
For example, for the set , the arithmetic mean, 25, is greater than the geometric mean, 18; AM-GM guarantees this is always the case.
The equality condition of this inequality states that the arithmetic mean and geometric mean are equal if and only if all members of the set are equal.
AM-GM can be used fairly frequently to solve Olympiad-level inequality problems, such as those on the USAMO and IMO.
Proof
There are so many proofs of AM-GM that they have an article to themselves: Proofs of AM-GM.
Weighted Form
The weighted form of AM-GM is given by using weighted averages. For example, the weighted arithmetic mean of and with is and the geometric is .
AM-GM applies to weighted averages. Specifically, the weighted AM-GM Inequality states that if are nonnegative real numbers, and are nonnegative real numbers (the "weights") which sum to 1, then or, in more compact notation, Equality holds if and only if for all integers such that and . We obtain the unweighted form of AM-GM by setting .
Extensions
- The power mean inequality is a generalization of AM-GM which places the arithemetic and geometric means on a continuum of different means.
- The root-square-mean arithmetic-mean geometric-mean harmonic-mean inequality is special case of the power mean inequality.
Problems
Introductory
Intermediate
- Find the minimum value of for .
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Olympiad
- Let , , and be positive real numbers. Prove that
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