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}.

If x1, x2, . . ., xn > 0, this is equal to the exponential of the arithmetic mean of the natural 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 nonnegative weights α1, α2, . . ., αn be given. Set \alpha = \alpha_1 + \alpha_2 + \cdots + \alpha_n . If α > 0, then the inequality

\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}}

holds with equality if and only if all the xk with αk > 0 are equal. Here the convention 00 = 1 is used.

If all αk = 1, this reduces to the above AM-GM inequality.

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 induction

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). It can also be proven using the rearrangement inequality. Considering length and required prerequisites, the proof by induction given below is probably the best recommendation for first reading.


With the arithmetic mean

\mu=\frac{\ x_1 + \cdots + x_n}n

of the non-negative real numbers x1,...,xn, the AM-GM statement is equivalent to

\mu^n\ge x_1 x_2 \cdots x_n\,

with equality if and only if μ = xi for all i = 1,...,n.


For the following proof we apply mathematical induction and only well-known rules of arithmetic.

Induction basis: For n = 1 the statement is true with equality.

Induction hypothesis: Suppose that the AM-GM statement holds for all choices of n non-negative real numbers.

Induction step: Consider n + 1 non-negative real numbers. Their arithmetic mean μ satisfies

(n+1)\mu=\ x_1 + \cdots + x_n + x_{n+1}.\,

If all numbers are equal to μ, then we have equality in the AM-GM statement and we are done. Otherwise we may find one number that is greater than μ and one that is smaller than μ, say xn > μ and xn+1 < μ. Then

(x_n-\mu)(\mu-x_{n+1})>0\,.\qquad(*)

Now consider the n numbers

x_1, \ldots, x_{n-1}, x_n'   with   x_n':=x_n+x_{n+1}-\mu\ge x_n-\mu>0\,,

which are also non-negative. Since

n\mu=x_1 + \cdots + x_{n-1} + \underbrace{x_n+x_{n+1}-\mu}_{=\,x_n'},

μ is also the arithmetic mean of x_1, \ldots, x_{n-1}, x_n' and the induction hypothesis implies

\mu^{n+1}=\mu^n\cdot\mu\ge x_1x_2 \cdots x_{n-1} x_n'\mu.\qquad(**)

Due to (*) we know that

(\underbrace{x_n+x_{n+1}-\mu}_{=\,x_n'})\mu-x_nx_{n+1}=(x_n-\mu)(\mu-x_{n+1})>0,

hence

x_n'\mu>x_nx_{n+1}\,,\qquad({*}{*}{*})

in particular μ > 0. Therefore, if at least one of the numbers x1,...,xn−1 is zero, then we already have strict inequality in (**). Otherwise the right-hand side of (**) is positive and strict inequality is obtained by using the estimate (***). Therefore, substituting (***) into (**) gives in both cases

\mu^{n+1}>x_1x_2 \cdots x_{n-1} x_nx_{n+1}\,,

which completes the proof.

[edit] Proof by Pólya

George Pólya provided the following proof, using the exponential function and the inequality ex ≥ 1 + x, which is valid for every real number x. To verify this inequality, note that both sides as well as their first derivatives agree for x = 0 and that the exponential function is strictly convex, because its second derivative is positive for every real x. For this reason, also the equality ex = 1 + x only holds for x = 0.

Let μ be the arithmetic mean, and let ρ be the geometric mean of x1, ..., xn. If all x1, ..., xn are equal, then μ = ρ.

It remains to prove the strict inequality μ > ρ if x1, ..., xn ≥ 0 are not all equal. Then, in particular, they are not all zero, hence μ > 0.

If we substitute xi /μ − 1 for x in the above inequality ex ≥ 1 + x we get that

\exp\Bigl({x_i \over \mu} - 1\Bigr)\ge {x_i \over \mu}\,

for each i and strict inequality for those i with xi ≠ μ. Since xi /μ ≥ 0, we can multiply all these inequalities together, side by side, for i = 1, ..., n, to obtain

\prod_{i=1}^n\exp\Bigl({x_i \over \mu} - 1\Bigr) > \prod_{i=1}^n {x_i \over \mu}\,,

where we get strict inequality because no factor on the left-hand side is zero and there was strict inequality for at least one i. Using the functional equation of the exponential function, we get

\exp\biggl(\frac1\mu\sum_{i=1}^n x_i - n\biggr) > \prod_{i=1}^n {x_i \over \mu}\,.\qquad(*)

Since μ is the arithmetic mean, the summation in the parentheses on the left of (*) can be reduced to

\sum_{i=1}^n x_i = n\mu\,.

Thus, the left-hand side of the inequality (*) is exp(n − n) = 1. Since ρ is the geometric mean, the product on the right of (*) can be rewritten as

\frac1{\mu^n}\prod_{i=1}^n x_i = {\rho^n \over \mu^n}.

So (*) reduces to 1 > ρn/μn and hence μ > ρ.

[edit] Proof by Cauchy

The following proof by cases relies directly on well-known rules of arithmetic but employs the rarely-used technique of forward-backward-induction. 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:

\begin{align} x_1 & \ne x_2 \\[3pt] x_1 - x_2 & \ne 0 \\[3pt] \left( x_1 - x_2 \right) ^2 & > 0 \\[3pt] x_1^2 - 2 x_1 x_2 + x_2^2 & > 0 \\[3pt] x_1^2 + 2 x_1 x_2 + x_2^2 & > 4 x_1 x_2 \\[3pt] \left( x_1 + x_2 \right) ^2& > 4 x_1 x_2 \\[3pt] \Bigl( \frac{x_1 + x_2}{2} \Bigr)^2 & > x_1 x_2 \\[3pt] \frac{x_1 + x_2}{2} & > \sqrt{x_1 x_2} \end{align}

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:

\begin{align} \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} \\[7pt] & \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} \\[7pt] & \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}}} \\[7pt] & = \sqrt[2^k]{x_1 \cdot x_2 \cdots x_{2^k}} \end{align}

where in the first inequality, the two sides are equal only 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 is 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:

\begin{align} \alpha & = \frac{x_1 + x_2 + \cdots + x_n}{n} \\[6pt] & = \frac{\frac{m}{n} \left( x_1 + x_2 + \cdots + x_n \right)}{m} \\[6pt] & = \frac{x_1 + x_2 + \cdots + x_n + \frac{m-n}{n} \left( x_1 + x_2 + \cdots + x_n \right)}{m} \\[6pt] & = \frac{x_1 + x_2 + \cdots + x_n + \left( m-n \right) \alpha}{m} \\[6pt] & = \frac{x_1 + x_2 + \cdots + x_n + x_{n+1} + \cdots + x_m}{m} \\[6pt] & > \sqrt[m]{x_1 \cdot x_2 \cdots x_n \cdot x_{n+1} \cdots x_m} \\[6pt] & = \sqrt[m]{x_1 \cdot x_2 \cdots x_n \cdot \alpha^{m-n}}\,, \end{align}

so

\begin{align} \alpha^m & > x_1 \cdot x_2 \cdots x_n \cdot \alpha^{m-n} \\[3pt] \alpha^n & > x_1 \cdot x_2 \cdots x_n \\[3pt] \alpha & > \sqrt[n]{x_1 \cdot x_2 \cdots x_n} \end{align}

as desired.

[edit] Proof of the generalized AM-GM inequality using Jensen's inequality

Using the finite form of Jensen's inequality for the natural logarithm, we can prove the inequality between the weighted arithmetic mean and the weighted geometric mean stated above.

Since an xk with weight αk = 0 has no influence on the inequality, we may assume in the following that all weights are positive. If all xk are equal, then equality holds. Therefore, it remains to prove strict inequality if they are not all equal, which we will assume in the following, too. If at least one xk is zero (but not all), then the weighted geometric mean is zero, while the weighted arithmetic mean is positive, hence strict inequality holds. Therefore, we may assume also that all xk are positive.

Since the natural logarithm is strictly concave, the finite form of Jensen's inequality and the functional equations of the natural logarithm imply

\ln\biggl(\frac{\alpha_1x_1+\cdots+\alpha_nx_n}\alpha\biggr) >\frac{\alpha_1}\alpha\ln x_1+\cdots+\frac{\alpha_n}\alpha\ln x_n =\ln \sqrt[\alpha]{x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}}.

Since the natural logarithm is strictly increasing,

\frac{\alpha_1x_1+\cdots+\alpha_nx_n}\alpha >\sqrt[\alpha]{x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}}.

[edit] References