Difference between revisions of "Arithmetic Mean-Geometric Mean Inequality"

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

Revision as of 20:12, 13 May 2020

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.


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 $a_1,a_2,\ldots,a_n$, the following always holds: \[\frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n}\] Using the shorthand notation for summations and products: \[\sum_{i=1}^{n}\frac{a_i}{n} \geq \prod\limits_{i=1}^{n}a_i^{\frac{1}{n}} .\] For example, for the set $\{9,12,54\}$, 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.


See here: 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 $x$ and $y$ with $3:1$ is $\frac{3x+1y}{3+1}$ and the geometric is $\sqrt[3+1]{x^3y}$.

AM-GM applies to weighted averages. Specifically, the weighted AM-GM Inequality states that if $a_1, a_2, \dotsc, a_n$ are nonnegative real numbers, and $\lambda_1, \lambda_2, \dotsc, \lambda_n$ are nonnegative real numbers (the "weights") which sum to 1, then \[\lambda_1 a_1 + \lambda_2 a_2 + \dotsb + \lambda_n a_n \ge a_1^{\lambda_1} a_2^{\lambda_2} \dotsm a_n^{\lambda_n},\] or, in more compact notation, \[\sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i} .\] Equality holds if and only if $a_i = a_j$ for all integers $i, j$ such that $\lambda_i \neq 0$ and $\lambda_j \neq 0$. We obtain the unweighted form of AM-GM by setting $\lambda_1 = \lambda_2 = \dotsb = \lambda_n = 1/n$.





  • Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$.



  • 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)

See Also

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