Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Arithmetic Mean-Geometric Mean Inequality"
(expand)
Line 1: Line 1:
−
The '''Arithmetic Mean-Geometric Mean Inequality''' ('''AM-GM''' or '''AMGM''') is an elementary inequality, generally one of the first ones taught in number theory courses.  
+
The '''Arithmetic Mean-Geometric Mean Inequality''' ('''AM-GM''' or '''AMGM''') is an elementary [[inequality]], generally one of the first ones taught in inequality courses.
  
−
== Inequality ==
+
== Theorem ==
−
The AM-GM states that for any multiset of positive real numbers, the arithmetic mean of the set is greater than or equal to the geometric mean of the set. Or:
+
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:
  
 
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:
  
 
<math>\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)\geq\sqrt[n]{a_1a_2\cdots a_n}</math>
 
<math>\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)\geq\sqrt[n]{a_1a_2\cdots a_n}</math>
 +
 +
Using the shorthand notation for [[summation]]s and [[product]]s:
 +
 +
<math>\left(\frac{\sum\limits_{i=1}^{n}a_i}{n}\right)\geq\sqrt[n]{\prod\limits_{i=1}^{n}a_i}</math>
  
 
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.   
  
−
The [[equality condition]] of this [[inequality]] states that the AM and GM are equal if and only if all members of the set are equal.
+
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.
  
−
The AM-GM inequalitiy is a specific case of the [[power mean inequality]].  Both 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]].
+
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]].
  
−
== Introductory Problems ==
+
=== Proof ===
−
* [[1983_AIME_Problems/Problem_9 | 1983 AIME Problem 9]]
+
{{incomplete|proof}}
 +
=== Weighted Form ===
 +
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.
  
−
== See also ==
+
==Extensions==
 +
*The [[power mean inequality]] is a useful inequality extending on the arithmetic mean side of the inequality.
 +
*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.
 +
==Problems==
 +
=== Introductory ===
 +
=== Intermediate ===
 +
* Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.
 +
([[1983 AIME Problems/Problem 9|Source]])
 +
=== Olympiad ===
 +
* Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers.  Prove that
 +
<center>
 +
<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]])
 +
== See Also ==
  
 
* [[RMS-AM-GM-HM]]
 
* [[RMS-AM-GM-HM]]
 
* [[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/2006/5/29.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]
 
* [http://www.mathideas.org/problems/2006/5/31.pdf Inequalities: An Application of RMS-AM-GM-HM by Adeel Khan]
  
−
{{wikify}}
 
  
−
{{stub}}
 
 
[[Category:Inequality]]
 
[[Category:Inequality]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 20:53, 20 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 positive real numbers, the arithmetic mean of the set is greater than or equal to the geometric mean of the set. Algebraically, this is written:

For a set of nonnegative real numbers $a_1,a_2,\ldots,a_n$, the following always holds:

$\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)\geq\sqrt[n]{a_1a_2\cdots a_n}$

Using the shorthand notation for summations and products:

$\left(\frac{\sum\limits_{i=1}^{n}a_i}{n}\right)\geq\sqrt[n]{\prod\limits_{i=1}^{n}a_i}$

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.

AMGM can be used fairly frequently to solve Olympiad-level inequality problems, such as those on the USAMO and IMO.

Proof

Template:Incomplete

Weighted Form

The weighted form of AMGM is given by using weighted averages. For example, the weighted arithmetic mean of $x$ and $y$ with $3:1$ is $\frac{3y+x}{2\cdot 3}$ and the geometric is $\sqrt[2+3]{xy^3}$. AMGM applies to weighted averages.

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Arithmetic_Mean-Geometric_Mean_Inequality&oldid=21425"