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.

Contents

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 %2B x_2 %2B \cdots %2B 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} %2B \ln {x_2} %2B \cdots %2B \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, . . ., xn,

\frac{x_1 %2B x_2 %2B \cdots %2B x_n}{n} \geq \sqrt[n]{x_1 \cdot x_2 \cdots x_n},

and that equality holds if and only if x1 = x2 = . . . = xn.

Geometric interpretation

In two dimensions, 2x1 + 2x2 is the perimeter of a rectangle with sides of length x1 and x2. Similarly, 4√(x1x2) is the perimeter of a square with the same area. Thus for n = 2 the AM-GM inequality states that a square has the smallest perimeter among all rectangles with equal area.

The full inequality is an extension of this idea to n dimensions. Every vertex on an n-dimensional box is connected to n edges. If these edges are given lengths x1, x2, ..., xn, then x1 + x2 + … + xn is the total length of edges connected to the vertex. There are 2n vertices, so multiply this by 2n; however, since each edge meets two vertices, that counts every edge twice. Thus we divide by 2 and conclude that there are n2n−1 edges. There are equally many edges of each length and n lengths; hence there are 2n−1 edges of each length. The total edge-length is therefore 2n−1(x1 + x2 + … + xn). On the other hand, 2^{n-1} n \sqrt[n]{x_1 x_2 \cdots x_n} is the total length of edges connected to a vertex on an n-dimensional cube of equal volume. Since the inequality says

\sqrt[n]{x_1\cdots x_n} \le {x_1 %2B \cdots %2B x_n \over n},

we get

n2^{n-1} \sqrt[n]{x_1\cdots x_n} \le 2^{n-1}(x_1 %2B \cdots %2B x_n). \,

Thus the AM–GM inequality states that an n-cube has the smallest sum of lengths of edges connected to each vertex among all n-dimensional boxes with the same volume. [1]

Generalizations

\frac{\alpha_1 x_1 %2B \alpha_2 x_2 %2B \cdots %2B \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 include these:

Example application

Consider the following function:

f(x,y,z) = \frac{x}{y} %2B \sqrt{\frac{y}{z}} %2B \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} %2B \frac{1}{2} \sqrt{\frac{y}{z}} %2B \frac{1}{2} \sqrt{\frac{y}{z}} %2B \frac{1}{3} \sqrt[3]{\frac{z}{x}} %2B \frac{1}{3} \sqrt[3]{\frac{z}{x}} %2B \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}}.

Proofs

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 %2B \cdots %2B 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%2B1)\mu=\ x_1 %2B \cdots %2B x_n %2B x_{n%2B1}.\,

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%2B1})>0\,.\qquad(*)

Now consider the n numbers

x_1, \ldots, x_{n-1}, x_n'   with   x_n':=x_n%2Bx_{n%2B1}-\mu\ge x_n-\mu>0\,,

which are also non-negative. Since

n\mu=x_1 %2B \cdots %2B x_{n-1} %2B \underbrace{x_n%2Bx_{n%2B1}-\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%2B1}=\mu^n\cdot\mu\ge x_1x_2 \cdots x_{n-1} x_n'\mu.\qquad(**)

Due to (*) we know that

(\underbrace{x_n%2Bx_{n%2B1}-\mu}_{=\,x_n'})\mu-x_nx_{n%2B1}=(x_n-\mu)(\mu-x_{n%2B1})>0,

hence

x_n'\mu>x_nx_{n%2B1}\,,\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 (***) to get a lower bound of the right-hand side of (**). Thus, in both cases we get

\mu^{n%2B1}>x_1x_2 \cdots x_{n-1} x_nx_{n%2B1}\,,

which completes the proof.

Proof by Pólya

George Pólya provided a proof similar to what follows. Let f(x) = ex−1 − x, with derivative f'(x) = ex−1 − 1. Observe f'(1) = 0 and hence that f has an absolute minimum of f(1) = 0. Now x ≤ ex−1 for all real x.

Consider a list of non-negative numbers a_1, a_2, \dots , a_n with arithmetic mean μ. By repeated application of the above inequality, we obtain the following:

{ \frac{a_1}{\mu} \frac{a_2}{\mu} \cdots \frac{a_n}{\mu} } \le { e^{\frac{a_1}{\mu} - 1} e^{\frac{a_2}{\mu} - 1} \cdots e^{\frac{a_n}{\mu} - 1} } = \exp \left ( \frac{a_1}{\mu} - 1 %2B \frac{a_2}{\mu} - 1 %2B \cdots %2B \frac{a_n}{\mu} - 1 \right ). \qquad (1)

But the exponential argument can be simplified:

\frac{a_1}{\mu} - 1 %2B \frac{a_2}{\mu} - 1 %2B \cdots %2B \frac{a_n}{\mu} - 1 = \frac{(a_1 %2B a_2 %2B \cdots %2B a_n)}{\mu} - n = n - n = 0.

Returning to (1),

\frac{a_1 a_2 \cdots a_n}{\mu^n} \le e^0 = 1,

which produces the result: [2]

a_1 a_2 \cdots a_n \le \mu^n \implies \sqrt[n]{a_1 a_2 \cdots a_n} \le \mu.

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.

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:

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

as desired.

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 %2B x_2 %2B \cdots %2B x_{2^k}}{2^k} & {} =\frac{\frac{x_1 %2B x_2 %2B \cdots %2B x_{2^{k-1}}}{2^{k-1}} %2B \frac{x_{2^{k-1} %2B 1} %2B x_{2^{k-1} %2B 2} %2B \cdots %2B x_{2^k}}{2^{k-1}}}{2} \\[7pt] & \ge \frac{\sqrt[2^{k-1}]{x_1 x_2 \cdots x_{2^{k-1}}} %2B \sqrt[2^{k-1}]{x_{2^{k-1} %2B 1} x_{2^{k-1} %2B 2} \cdots x_{2^k}}}{2} \\[7pt] & \ge \sqrt{\sqrt[2^{k-1}]{x_1 x_2 \cdots x_{2^{k-1}}} \sqrt[2^{k-1}]{x_{2^{k-1} %2B 1} x_{2^{k-1} %2B 2} \cdots x_{2^k}}} \\[7pt] & = \sqrt[2^k]{x_1 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}%2B1} = x_{2^{k-1}%2B2} = \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 %2B x_2 %2B \cdots %2B x_{2^k}}{2^k} > \sqrt[2^k]{x_1 x_2 \cdots x_{2^k}}

as desired.

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%2B1} = x_{n%2B2} = \cdots = x_m = \alpha.

We then have:

\begin{align} \alpha & = \frac{x_1 %2B x_2 %2B \cdots %2B x_n}{n} \\[6pt] & = \frac{\frac{m}{n} \left( x_1 %2B x_2 %2B \cdots %2B x_n \right)}{m} \\[6pt] & = \frac{x_1 %2B x_2 %2B \cdots %2B x_n %2B \frac{m-n}{n} \left( x_1 %2B x_2 %2B \cdots %2B x_n \right)}{m} \\[6pt] & = \frac{x_1 %2B x_2 %2B \cdots %2B x_n %2B \left( m-n \right) \alpha}{m} \\[6pt] & = \frac{x_1 %2B x_2 %2B \cdots %2B x_n %2B x_{n%2B1} %2B \cdots %2B x_m}{m} \\[6pt] & > \sqrt[m]{x_1 x_2 \cdots x_n x_{n%2B1} \cdots x_m} \\[6pt] & = \sqrt[m]{x_1 x_2 \cdots x_n \alpha^{m-n}}\,, \end{align}

so

\begin{align} \alpha^m & > x_1 x_2 \cdots x_n \alpha^{m-n} \\[5pt] \alpha^n & > x_1 x_2 \cdots x_n \\[5pt] \alpha & > \sqrt[n]{x_1 x_2 \cdots x_n} \end{align}

as desired.

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%2B\cdots%2B\alpha_nx_n}\alpha\biggr) >\frac{\alpha_1}\alpha\ln x_1%2B\cdots%2B\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%2B\cdots%2B\alpha_nx_n}\alpha >\sqrt[\alpha]{x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}}.

See also

References

  1. ^ Steele, J. Michael (2004). The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities. MAA Problem Books Series. Cambridge University Press. ISBN 978-0521546775. OCLC 54079548. 
  2. ^ Arnold, Denise; Arnold, Graham (1993). Four unit mathematics. Hodder Arnold H&S. p. 242. ISBN 978-0340543351. OCLC 38328013.