Inequality of arithmetic and geometric means

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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.

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[edit] Background

The arithmetic mean, or less precisely the average, of a list of n numbers x1, x2, . . ., xn 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:

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

[edit] The inequality

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

\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 = . . . = xn,

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

[edit] Generalizations

There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers x1, x2, . . ., xn and the positive weights α1, α2, . . ., αn be given. Set \alpha = \alpha_1 + \alpha_2 + \cdots + \alpha_n. Then the following inequality holds

\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 xk are equal.

Other generalizations of the inequality of arithmetic and geometric means are given by Muirhead's inequality, MacLaurin's inequality, and the generalized mean inequality.

[edit] 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}}.

[edit] 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.

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.

[edit] 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.

[edit] 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.

[edit] 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.

[edit] The subcase where n = 2k

Consider the case where n = 2k, where k is a positive integer. We proceed by mathematical induction.

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 = 2k−1, and we wish to show that it holds for n = 2k. 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 2k 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.

[edit] The subcase where n < 2k

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, . . ., 2k, . . . 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.

[edit] 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.

Let μ be the arithmetic mean, and let ρ be the geometric mean. If we substitute xi /μ − 1 for x in the above inequality we get:

\exp\left({x_i \over \mu} - 1\right)\geq {x_i \over \mu}\mbox{ for each }i.\,

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

\exp\left(\sum_i {x_i \over \mu} - n\right)\geq \prod_i \left({x_i \over \mu}\right).

The summation in the parentheses 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.

[edit] 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\,

if

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 xm and xn which are not equal to one. We express them as

x_a = 1 + \alpha\,
x_b = 1 - \beta.\,

their sum

xa + xb = 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 xa xb 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

[edit] Proof using Jensen's inequality

Another way to show AM-GM is to apply Jensen's inequality to the logarithm function, which yields:

\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.

[edit] References

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