Inequality of arithmetic and geometric means

In mathematics, the arithmetic mean of numbers x₁, ..., x_n is just what pupils are taught to call the average, i.e., it is

$\overline{x}=(x_1+\cdots+x_n)/n.$

The geometric mean is to multiplication as the arithmetic mean is to addition. Just as adding n terms all equal to the arithmetic mean yields the sum x₁ + ... + x_n, so multiplying n factors all equal to the geometric mean yields the product x₁ ... x_n (these n numbers must be non-negative). In other words, the geometric mean is

$(x_1\cdots x_n)^{1/n}.$

This is the same as taking the logarithm of each of the numbers x₁, ..., x_n, finding the ordinary average of those logarithms, and then taking the antilogarithm of the resulting average. (The base of the logarithm and that of the antilogarithm must both be the same; beyond that it makes no difference which base is used.)

Contents

1 The inequality

2 Proofs

3 Example application

4 Generalization

The inequality

The inequality that is our topic states simply that the geometric mean of a list of non-negative numbers is always less than or equal to their arithmetic mean, and the two means are equal if and only if the n numbers whose mean is taken are all equal to each other. In mathematical notation, if x₁, ..., x_n ≥ 0, then

$(x_1\cdots x_n)^{1/n}\leq{x_1+\cdots+x_n \over n},$

and equality holds if and only if

$x_1=\cdots=x_n.$

Proofs

One way to prove this inequality is by inferring it as a corollary of Jensen's inequality. There are many other derivations.

For the case of just two numbers, a, b > 0 the statement of AM-GM is

$\frac{a+b}{2} \geq \sqrt{ab}.$

We can prove this by noting:

$(a-b)^{2} \geq 0$

by the fact that squares are non-negative. Now by expanding and adding 4ab to both sides, we get

$a^2+2ab+b^2 \geq 4ab.$

By taking square roots of both sides and then dividing by 2 we get

$\frac{a+b}{2} \geq \sqrt{ab}.$

The general form of the AM-GM inequality,

$\frac{1}{n}\sum_{k=1}^n a_{k} \geq \sqrt[n]{\prod_{j=1}^{n} a_{j}}$

with equality if and only if

$a_1 = a_2 = \cdots = a_n$

for a₁, ..., a_n ≥ 0, can be proved by mathematical induction.

Example application

$\frac{x}{y} + \sqrt{\frac{y}{z}} + \sqrt[3]{\frac{z}{x}} = \frac{x}{y} + \frac{1}{2}\sqrt{\frac{y}{z}} + \frac{1}{2}\sqrt{\frac{y}{z}} + \frac{1}{3}\sqrt[3]{\frac{z}{x}} + \frac{1}{3}\sqrt[3]{\frac{z}{x}} + \frac{1}{3}\sqrt[3]{\frac{z}{x}}$

Applying the weighted AM-GM inequality to the six terms on the right hand side, we have

$\frac{1}{6}\left(\frac{x}{y} + \frac{1}{2}\sqrt{\frac{y}{z}} + \frac{1}{2}\sqrt{\frac{y}{z}} + \frac{1}{3}\sqrt[3]{\frac{z}{x}} + \frac{1}{3}\sqrt[3]{\frac{z}{x}} + \frac{1}{3}\sqrt[3]{\frac{z}{x}}\right) \geq \sqrt[6]\frac{1}{2\cdot 2\cdot 3\cdot 3\cdot 3}.$

Note that the right hand side is the arithmetic means of the six terms, while the left hand side is the geometric mean. Multiplying both sides by 6 and subsituting the first expression in, we have

$\frac{x}{y} + \sqrt{\frac{y}{z}} + \sqrt[3]{\frac{z}{x}} \geq \sqrt[6]{2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3} = 2^{\frac{2}{3}}\cdot 3^{\frac{1}{2}}.$

The equality condition, or minimum, is guaranteed by AM-GM when

$\frac{1}{3}\sqrt[3]{\frac{z}{x}} = \frac{1}{2}\sqrt{\frac{y}{z}} = \frac{x}{y}.$

Generalization

This inequality is a special case of Muirhead's inequality.

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