NationMaster - Encyclopedia: Inequality of arithmetic and geometric means

In mathematics, the inequality of arithmetic and geometric means, or more briefly the AM-GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is the same. Euclid, Greek mathematician, 3rd century BC, known today as the father of geometry; shown here in a detail of The School of Athens by Raphael. ... In mathematics and statistics, the arithmetic mean (or simply the mean) of a list of numbers is the sum of all the members of the list divided by the number of items in the list. ... In mathematics, the real numbers may be described informally in several different ways. ... The geometric mean of a set of positive data is defined as the nth root of the product of all the members of the set, where n is the number of members. ...

1 Background
2 The inequality
3 Generalizations
4 Example application
5 Proof by Cauchy
- 5.1 The case where all the terms are equal
- 5.2 The case where not all the terms are equal
6 Proof by Pólya
7 Proof by induction
8 Proof using Jensen's inequality
9 References

Background

The arithmetic mean, or less precisely the average, of a list of n numbers x₁, x₂, . . ., x_n is the sum of the numbers divided by n:

$frac{x_1 + x_2 + cdots + x_n}{n}$ .

The geometric mean is similar, except that it is only defined for a list of nonnegative real numbers, and uses multiplication and a root in place of addition and division:

$sqrt[n]{x_1 cdot x_2 cdots x_n}.$

This is equivalent to the antilogarithm of the arithmetic mean of the list of logarithms of the numbers: In mathematics, if two variables of bn = x are known, the third can be found. ... Logarithms to various bases: is to base e, is to base 10, and is to base 1. ...

$exp left( frac{ln {x_1} + ln {x_2} + cdots + ln {x_n}}{n} right).$

The inequality

Restating the inequality using mathematical notation, we have that for any list of n nonnegative real numbers x₁, x₂, . . ., x_n,

$frac{x_1 + x_2 + cdots + x_n}{n} geq sqrt[n]{x_1 cdot x_2 cdots x_n}$ ,

and that if and only if x₁ = x₂ = . . . = x_n,

$frac{x_1 + x_2 + cdots + x_n}{n} = sqrt[n]{x_1 cdot x_2 cdots x_n}$ .

Generalizations

There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers x₁, x₂, . . ., x_n and the positive weights α₁, α₂, . . ., α_n be given. Set $alpha = alpha_1 + alpha_2 + cdots + alpha_n$ . Then the following inequality holds In statistics, given a set of data, X = { x1, x2, ..., xn} and corresponding weights, W = { w1, w2, ..., wn} the weighted mean is calculated as Note that if all the weights are equal, the weighted mean is the same as the arithmetic mean. ... In statistics, given a set of data, X = { x1, x2, ..., xn} and corresponding weights, W = { w1, w2, ..., wn} the weighted geometric mean is calculated as Note that if all the weights are equal, the weighted geometric mean is the same as the geometric mean. ...

$frac{alpha_1 x_1 + alpha_2 x_2 + cdots + alpha_n x_n}{alpha} geq sqrt[alpha]{x_1^{alpha_1} x_2^{alpha_2} cdots x_n^{alpha_n}}$ ,

with equality if and only if all the $x k$ are equal.

Other generalizations of the inequality of arithmetic and geometric means are given by Muirhead's inequality and Generalized mean inequality. In mathematics, Muirheads inequality, also known as the bunching method, generalizes the inequality of arithmetic and geometric means. ... A generalized mean, also known as power mean or HÃ¶lder mean, is an abstraction of the arithmetic, geometric and harmonic means. ...

Example application

Consider the following function:

$f(x,y,z) = frac{x}{y} + sqrt{frac{y}{z}} + sqrt[3]{frac{z}{x}}$

for x, y, and z all positive real numbers. Suppose we wish to find the minimum value of this function. Rewriting a bit, and applying the AM-GM inequality, we have:

$f(x,y,z),;$	$= 6 cdot frac{ frac{x}{y} + frac{1}{2} sqrt{frac{y}{z}} + frac{1}{2} sqrt{frac{y}{z}} + frac{1}{3} sqrt[3]{frac{z}{x}} + frac{1}{3} sqrt[3]{frac{z}{x}} + frac{1}{3} sqrt[3]{frac{z}{x}} }{6}$

	$ge 6 cdot sqrt[6]{ frac{x}{y} cdot frac{1}{2} sqrt{frac{y}{z}} cdot frac{1}{2} sqrt{frac{y}{z}} cdot frac{1}{3} sqrt[3]{frac{z}{x}} cdot frac{1}{3} sqrt[3]{frac{z}{x}} cdot frac{1}{3} sqrt[3]{frac{z}{x}} }$

	$= 6 cdot sqrt[6]{ frac{1}{2 cdot 2 cdot 3 cdot 3 cdot 3} frac{x}{y} frac{y}{z} frac{z}{x} }$

	$= 2^{2/3} cdot 3^{1/2}$

Further, we know that the two sides are equal exactly when all the terms of the mean are equal:

$f(x,y,z) = 2^{2/3} cdot 3^{1/2} quad mbox{when} quad frac{x}{y} = frac{1}{2} sqrt{frac{y}{z}} = frac{1}{3} sqrt[3]{frac{z}{x}}.$

Proof by Cauchy

There are several ways to prove the AM-GM inequality; for example, it can be inferred from Jensen's inequality, using the concave function ln(x).[1] It can also be proven using the rearrangement inequality. In mathematics, Jensens inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function. ... Let be real numbers and be any permutation of . ...

The following proof by cases relies directly on well-known rules of arithmetic. It is essentially from Augustin Louis Cauchy and can be found in his Cours d'analyse. Augustin Louis Cauchy Augustin Louis Cauchy (August 21, 1789 â€“ May 23, 1857) was a French mathematician. ...

The case where all the terms are equal

If all the terms are equal:

$x_1 = x_2 = cdots = x_n$

then their sum is nx₁, so their arithmetic mean is x₁; and their product is x₁ⁿ, so their geometric mean is x₁; therefore, the arithmetic mean and geometric mean are equal, as desired.

The case where not all the terms are equal

It remains to show that if not all the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible when n > 1.

This case is significantly more complex, and we divide it into subcases.

The subcase where n = 2

If n = 2, then we have two terms, x₁ and x₂, and since (by our assumption) not all terms are equal, we have:

$x_1,;$	$ne x_2,;$
$x_1 - x_2,;$	$ne 0,;$
$left( x_1 - x_2 right) ^2,;$	$> 0,;$
$x_1^2 - 2 x_1 x_2 + x_2^2,;$	$> 0,;$
$x_1^2 + 2 x_1 x_2 + x_2^2,;$	$> 4 x_1 x_2,;$
$left( x_1 + x_2 right) ^2,;$	$> 4 x_1 x_2,;$
$left( frac{x_1 + x_2}{2} right)^2,;$	$> x_1 x_2,;$
$frac{x_1 + x_2}{2},;$	$> sqrt{x_1 x_2},;$

as desired.

The subcase where n = 2^k

Consider the case where n = 2^k, where k is a positive integer. We proceed by mathematical induction. Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers. ...

In the base case, k = 1, so n = 2. We have already shown that the inequality holds where n = 2, so we are done.

Now, suppose that for a given k > 1, we have already shown that the inequality holds for n = 2^k−1, and we wish to show that it holds for n = 2^k. To do so, we proceed as follows:

$frac{x_1 + x_2 + cdots + x_{2^k}}{2^k}$	$=frac{frac{x_1 + x_2 + cdots + x_{2^{k-1}}}{2^{k-1}} + frac{x_{2^{k-1} + 1} + x_{2^{k-1} + 2} + cdots + x_{2^k}}{2^{k-1}}}{2}$

	$ge frac{sqrt[2^{k-1}]{x_1 cdot x_2 cdots x_{2^{k-1}}} + sqrt[2^{k-1}]{x_{2^{k-1} + 1} cdot x_{2^{k-1} + 2} cdots x_{2^k}}}{2}$

	$ge sqrt{sqrt[2^{k-1}]{x_1 cdot x_2 cdots x_{2^{k-1}}} cdot sqrt[2^{k-1}]{x_{2^{k-1} + 1} cdot x_{2^{k-1} + 2} cdots x_{2^k}}}$

	$= sqrt[2^k]{x_1 cdot x_2 cdots x_{2^k}}$

where in the first inequality, the two sides are only equal if both of the following are true:

$x_1 = x_2 = cdots = x_{2^{k-1}}$

$x_{2^{k-1}+1} = x_{2^{k-1}+2} = cdots = x_{2^k}$

(in which case the first arithmetic mean and first geometric mean are both equal to x₁, and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all 2^k numbers are equal, it's not possible for both inequalities to be equalities, so we know that:

$frac{x_1 + x_2 + cdots + x_{2^k}}{2^k} > sqrt[2^k]{x_1 cdot x_2 cdots x_{2^k}}$

as desired.

The subcase where n < 2^k

If n is not a natural power of 2, then it is certainly less than some natural power of 2, since the sequence 2, 4, 8, . . ., 2^k, . . . is unbounded above. Therefore, without loss of generality, let m be some natural power of 2 that is greater than n.

So, if we have n terms, then let us denote their arithmetic mean by α, and expand our list of terms thus:

$x_{n+1} = x_{n+2} = cdots = x_m = alpha.$

We then have:

$alpha,;$	$= frac{x_1 + x_2 + cdots + x_n}{n}$

	$= frac{frac{m}{n} left( x_1 + x_2 + cdots + x_n right)}{m}$

	$= frac{x_1 + x_2 + cdots + x_n + frac{m-n}{n} left( x_1 + x_2 + cdots + x_n right)}{m}$

	$= frac{x_1 + x_2 + cdots + x_n + left( m-n right) alpha}{m}$

	$= frac{x_1 + x_2 + cdots + x_n + x_{n+1} + cdots + x_m}{m}$

	$> sqrt[m]{x_1 cdot x_2 cdots x_n cdot x_{n+1} cdots x_m}$

	$= sqrt[m]{x_1 cdot x_2 cdots x_n cdot alpha^{m-n}},$
so
$alpha^m,;$	$> x_1 cdot x_2 cdots x_n cdot alpha^{m-n}$
$alpha^n,;$	$> x_1 cdot x_2 cdots x_n$
$alpha,;$	$> sqrt[n]{x_1 cdot x_2 cdots x_n}.$

as desired.

Proof by Pólya

George Pólya provided the following proof, using the inequality $e^x geq 1+x$ . The fact that all numbers are greater than or equal to zero allows us to manipulate inequalities without changing their directions. George PÃ³lya (December 13, 1887 â€“ September 7, 1985, in Hungarian PÃ³lya GyÃ¶rgy) was a Hungarian mathematician. ...

Let μ be the arithmetic mean, and let ρ be the geometric mean.

Let x = x_i/μ − 1. Using the above inequality:

$expleft({x_i over mu} - 1right)geq {x_i over mu}mbox{ for each }i.,$

Multiply all these inequalities together, side by side, for i = 1, ..., n, we have

$expleft(sum_i {x_i over mu} - nright)geq prod_i left({x_i over mu}right).$

The summation on the left can be reduced to

$sum_i x_i/mu = frac{sum_i x_i}{mu} = frac{sum_i x_i}{left(frac{sum_i x_i}{n}right)} = n.$

Thus, the left hand side of the inequality is exp(n − n) = 1.

The product on the right can be rewritten as

$frac{prod_i x_i}{mu^n} = {rho^n over mu^n}.$

So we have $1 geq rho^n / mu^n$ and hence $mu geq rho$ .

Proof by induction

It is easy to show that the AM-GM statement is equal to

$x_1 cdot x_2 cdots x_n le 1,$

$x_1 + cdots + x_n = n.,$

It's easy to show this by multiplying the AM-GM inequality by an appropriate value p.

$p = frac{n}{x_1 + cdots + x_n}$

Now it is easy to apply induction. Suppose that the inequality holds for k variables. If among k + 1 variables all are equal we are done. Otherwise we may find two variables x_m and x_n which are not equal to one. We express them as

$x_a = 1 + alpha,$

$x_b = 1 - beta.,$

their sum

x a + x b = 2 + α - β

substituting this expression in the general statement:

$x_1 + cdots + x_{m-1} + x_{m+1} + cdots + x_{n-1} + x_{n+1} + cdots + x_k + (x_a + x_b ) = k + 1$

$x_1 + cdots + x_{m-1} + x_{m+1} + cdots + x_{n-1} + x_{n+1} + cdots + x_k + (1+alpha - beta) = k$

now we are done if we show that the product x_a x_b is less than one. But it's obviously true (add proof), so

$x_1 cdots x_{k+1} le 1$

by dividing the expression by p we return to the classical representation of the AM-GM

Proof using Jensen's inequality

Another way to show AM-GM is to apply Jensen's inequality to the logarithm function, which yields: In mathematics, Jensens inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function. ... Logarithms to various bases: is to base e, is to base 10, and is to base 1. ...

$frac {log{x_1} + cdots + log{x_n}}{n} le log { frac{x_1 + cdots + x_n}{n} }.$

The inequality of arithmetic and geometric means follows by straightforward algebraic manipulations.

References

Augustin-Louis Cauchy, Cours d'analyse de l'École Royale Polytechnique, premier partie, Analyse algébrique, Paris, 1821. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.

Category: Inequalities

Results from FactBites:

PlanetMath: arithmetic-geometric-harmonic means inequality (158 words)
	See Also: arithmetic mean, geometric mean, harmonic mean, general means inequality, weighted power mean, power mean, root-mean-square, proof of general means inequality, Jensen's inequality, derivation of geometric mean as the limit of the power mean, minimal and maximal number, proof of arithmetic-geometric means inequality using Lagrange multipliers
	inequality, mean, arithmetic mean, geometric mean, harmonic mean
	This is version 5 of arithmetic-geometric-harmonic means inequality, born on 2001-08-18, modified 2004-06-05.

mean: Definition, Synonyms and Much More from Answers.com (3122 words)
	The mean is the arithmetic average of a set of values, or distribution; however, for skewed distributions, the mean is not necessarily the same as the middle value (median), or most likely (mode).
	The geometric mean is an average that is useful for sets of numbers that are interpreted according to their product and not their sum (as is the case with the arithmetic mean).
	The generalized mean, also known as the power mean or Hölder mean, is an abstraction of the arithmetic, geometric and harmonic means.

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Contents

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The inequality

Generalizations

Example application

Proof by Cauchy

The case where all the terms are equal

The case where not all the terms are equal

The subcase where n = 2

The subcase where n = 2k

The subcase where n < 2k

Proof by Pólya

Proof by induction

Proof using Jensen's inequality

References

Background

The subcase where n = 2^k

The subcase where n < 2^k