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

m (→‎Theorem)
(cleaned LaTeX code, elaborated in weighted section and corrected blatant error; made link to article of proofs)
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== Theorem ==
 
== Theorem ==
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The AM-GM states that for any [[set]] of [[positive]] [[real number]]s, the [[arithmetic mean]] of the set is greater than or [[equal]] to the [[geometric mean]] of the set. Algebraically, this is written:
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The 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.
  
 
For a set of nonnegative real numbers <math>a_1,a_2,\ldots,a_n</math>, the following always holds:
 
For a set of nonnegative real numbers <math>a_1,a_2,\ldots,a_n</math>, the following always holds:
 +
<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:
 +
<cmath> \sum_{i=1}^{n}a_i}/n \geq \prod\limits_{i=1}^{n}a_i^{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. 
  
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<center><math>\frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n}</math></center>
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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.
  
−
Using the shorthand notation for [[summation]]s and [[product]]s:
+
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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<center><math>\frac{\sum\limits_{i=1}^{n}a_i}{n}\geq\sqrt[n]{\prod\limits_{i=1}^{n}a_i}</math></center>
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=== Proof ===
  
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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.
+
There are so many proofs of AM-GM that they have an article to themselves:  [[Proofs of AM-GM]].
  
−
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.
+
=== Weighted Form ===
  
−
AMGM 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]].
+
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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=== Proof ===
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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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{{incomplete|proof}}
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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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=== Weighted Form ===
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or, in more compact notation,
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The weighted form of AMGM 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{3y+x}{2\cdot 3}</math> and the geometric is <math>\sqrt[2+3]{xy^3}</math>. AMGM applies to weighted averages.
+
<cmath> \sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i} . </cmath>
 +
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>.
 +
We obtain the unweighted form of AM-GM by setting <math>\lambda_1 = \lambda_2 = \dotsb = \lambda_n = 1/n</math>.
  
 
==Extensions==
 
==Extensions==
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*The [[power mean inequality]] is a useful inequality extending on the arithmetic mean side of the inequality.
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*The [[root-square-mean arithmetic-mean geometric-mean harmonic-mean inequality]] is an extension on both sides of the inequality with further types of means.
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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 [[root-square-mean arithmetic-mean geometric-mean harmonic-mean inequality]] is special case of the power mean inequality.
 +
 
 
==Problems==
 
==Problems==
 +
 
=== Introductory ===
 
=== Introductory ===
 
=== Intermediate ===
 
=== Intermediate ===
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([[1983 AIME Problems/Problem 9|Source]])
 
([[1983 AIME Problems/Problem 9|Source]])
 
=== Olympiad ===
 
=== Olympiad ===
 +
 
* Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers.  Prove that
 
* Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers.  Prove that
−
<center>
+
<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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<math>
 
−
(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3
 
−
</math>.
 
−
</center>
 
 
([[2004 USAMO Problems/Problem 5|Source]])
 
([[2004 USAMO Problems/Problem 5|Source]])
 +
 
== See Also ==
 
== See Also ==
  

Revision as of 17:34, 31 December 2007

The Arithmetic Mean-Geometric Mean Inequality (AM-GM or AMGM) is an elementary inequality, generally one of the first ones taught in inequality courses.

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 $a_1,a_2,\ldots,a_n$, 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: 

\[\sum_{i=1}^{n}a_i}/n \geq \prod\limits_{i=1}^{n}a_i^{1/n} .\] (Error compiling LaTeX. Unknown error_msg)

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

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 $x$ 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

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

External Links

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