Difference between revisions of "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. | 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. | ||
Revision as of 11:10, 5 July 2008
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
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:
![\[\frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n}\]](http://latex.artofproblemsolving.com/7/c/0/7c0baf73378c745491485cb635305a5f1ec5e70c.png)
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. Unknown error_msg)
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 ![$\sqrt[3+1]{x^3y}$](http://latex.artofproblemsolving.com/e/1/a/e1a3c8b1669b8b207f14189b7d4741a996cbda39.png)
.
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
![\[\lambda_1 a_1 + \lambda_2 a_2 + \dotsb + \lambda_n a_n \ge a_1^{\lambda_1} a_2^{\lambda_2} \dotsm a_n^{\lambda_n},\]](http://latex.artofproblemsolving.com/e/0/d/e0dff057770ac8a13ebae1829ed9490dd23337ee.png)
or, in more compact notation,
![\[\sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i} .\]](http://latex.artofproblemsolving.com/d/d/b/ddb44660822034c115f2c6a90c46fd7d3a78d0bc.png)
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.
- 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.
Problems
Introductory
Intermediate
- Find the minimum value of
for 
.
for 
.(Source)
Olympiad
- Let
, 
, and 
be positive real numbers. Prove that
, 
, and 
be positive real numbers. Prove that![\[(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3 .\]](http://latex.artofproblemsolving.com/a/c/3/ac39e58f37f4f105cad843d0dc44370c62f9440a.png)
(Source)
