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

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<cmath>  \frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n} </cmath>
 
<cmath>  \frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n} </cmath>
 
Using the shorthand notation for [[summation]]s and [[product]]s:
 
Using the shorthand notation for [[summation]]s and [[product]]s:
<cmath> \sum_{i=1}^{n}a_i}/n \geq \prod\limits_{i=1}^{n}a_i^{1/n} . </cmath>
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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>
 
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.   
 
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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=== Proof ===
 
=== Proof ===
  
There are so many proofs of AM-GM that they have an article to themselves:  [[Proofs of AM-GM]].
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See here:  [[Proofs of AM-GM]].
  
 
=== Weighted Form ===
 
=== Weighted Form ===
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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.
 
* 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.
 
* 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==
 
==Problems==
  
 
=== Introductory ===
 
=== Introductory ===
* 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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* 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|Solution]])
  
 
=== Intermediate ===
 
=== Intermediate ===
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* [[Algebra]]
 
* [[Algebra]]
 
* [[Inequalities]]
 
* [[Inequalities]]
 
==External Links==
 
* [http://www.mathideas.org/problems/2006/5/29.pdf Basic Inequalities by Adeel Khan]
 
* [http://www.mathideas.org/problems/2006/5/31.pdf Inequalities: An Application of RMS-AM-GM-HM by Adeel Khan]
 
  
  
 
[[Category:Inequality]]
 
[[Category:Inequality]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Latest revision as of 20:17, 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.

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 $a_1,a_2,\ldots,a_n$, 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.

Proof

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 $\lambda_1 = \lambda_2 = \dotsb = \lambda_n = 1/n$.

Extensions

Problems

Introductory

  • For nonnegative real numbers $a_1,a_2,\cdots a_n$, demonstrate that if $a_1a_2\cdots a_n=1$ then $a_1+a_2+\cdots +a_n\ge n$.
  • Find the maximum of $2 - a - \frac{1}{2a}$ for all positive $a$. (Solution)

Intermediate

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

(Source)

Olympiad

  • 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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