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Difference between revisions of "AM-GM Inequality"
(Too lazy to finesse problems right now, copy and pasted the problems section from the old article to make pretty one day.)
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The '''Power Mean Inequality''' relates all the different power means of a list of nonnegative reals. The power mean <math>M(p)</math> is defined as follows: <cmath>M(p) = \begin{cases} \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n}\right)^\frac{1}{p} &\text{if } p \neq 0 \\ \sqrt[n]{x_1 x_2 \cdots x_n} &\text{if } p = 0. \end{cases}</cmath> The Power Mean inequality then states that if <math>a>b</math>, then <math>M(a) \geq M(b)</math>, with equality holding if and only if <math>x_1 = x_2 = \cdots = x_n.</math> Plugging <math>p=1, 0</math> into this inequality reduces it to AM-GM, and <math>p=2, 1, 0, -1</math> gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
 
The '''Power Mean Inequality''' relates all the different power means of a list of nonnegative reals. The power mean <math>M(p)</math> is defined as follows: <cmath>M(p) = \begin{cases} \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n}\right)^\frac{1}{p} &\text{if } p \neq 0 \\ \sqrt[n]{x_1 x_2 \cdots x_n} &\text{if } p = 0. \end{cases}</cmath> The Power Mean inequality then states that if <math>a>b</math>, then <math>M(a) \geq M(b)</math>, with equality holding if and only if <math>x_1 = x_2 = \cdots = x_n.</math> Plugging <math>p=1, 0</math> into this inequality reduces it to AM-GM, and <math>p=2, 1, 0, -1</math> gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
  
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== Problems ==
  
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== Introductory examples ==
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=== Introductory ===
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WIP
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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>. ([[Solution to AM - GM Introductory Problem 1|Solution]])
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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]])
  
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== Intermediate examples ==
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=== Intermediate ===
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WIP
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* Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.
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([[1983 AIME Problems/Problem 9|Source]])
  
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== Olympiad examples ==
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=== Olympiad ===
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WIP
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* Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers.  Prove that
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<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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== More Problems ==
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([[2004 USAMO Problems/Problem 5|Source]])
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WIP
 
  
 
== See Also ==
 
== See Also ==
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* [[Proofs of AM-GM]]
 
* [[Mean Inequality Chain]]
 
* [[Mean Inequality Chain]]
 
* [[Power Mean Inequality]]
 
* [[Power Mean Inequality]]
 
* [[Cauchy-Schwarz Inequality]]
 
* [[Cauchy-Schwarz Inequality]]
 
* [[Inequality]]
 
* [[Inequality]]
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[[Category:Algebra]]
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[[Category:Inequalities]]
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[[Category:Definition]]

Revision as of 16:50, 29 December 2021

In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean; furthermore, the two means are equal if and only if every number in the list is the same.

In symbols, the inequality states that for any real numbers $x_1, x_2, \ldots, x_n \geq 0$, \[\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}\] with equality if and only if $x_1 = x_2 = \cdots = x_n$.

NOTE: This article is a work-in-progress and meant to replace the Arithmetic mean-geometric mean inequality article, which is of poor quality.

Proofs

Main article: Proofs of AM-GM

All known proofs of AM-GM use either induction or other, more advanced inequalities. Its proof is far more complicated than its usage in introductory competitions; consequentially, learning it is not recommended to students new to proofs. The most elementary proof of AM-GM utilizes Cauchy Induction, a variant of induction that involves proving a result for two, then using induction to prove it for all powers of two, and then a backward step where $n$ implies $n-1$.

Generalizations

The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.

Weighted AM-GM Inequality

The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights $\omega_1, \omega_2, \ldots, \omega_n \geq 0$ such that $\omega_1 + \omega_2 + \cdots + \omega_n = \omega$, \[\frac{\omega_1 x_1 + \omega_2 x_2 + \cdots + \omega_n x_n}{\omega} \geq \sqrt[\omega]{x_1^{\omega_1} x_2^{\omega_2} \cdots x_n^{\omega_n}},\] with equality if and only if $x_1 = x_2 = \cdots = x_n$. When $\omega_1 = \omega_2 = \cdots = \omega_n = 1/n$, the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.

Mean Inequality Chain

Main article: Mean Inequality Chain

The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that \[\sqrt{\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}} \geq \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} \geq \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}},\] with equality if and only if $x_1 = x_2 = \cdots = x_n$. As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.

Power Mean Inequality

Main article: Power Mean Inequality

The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean $M(p)$ is defined as follows: \[M(p) = \begin{cases} \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n}\right)^\frac{1}{p} &\text{if } p \neq 0 \\ \sqrt[n]{x_1 x_2 \cdots x_n} &\text{if } p = 0. \end{cases}\] The Power Mean inequality then states that if $a>b$, then $M(a) \geq M(b)$, with equality holding if and only if $x_1 = x_2 = \cdots = x_n.$ Plugging $p=1, 0$ into this inequality reduces it to AM-GM, and $p=2, 1, 0, -1$ gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.

Problems

Introductory

  • For nonnegative real numbers $a_1,a_2,\cdots a_n$, demonstrate that if $a_1a_2\cdots a_n=1$ then $a_1+a_2+\cdots +a_n\ge n$. (Solution)
  • 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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