Proofs of the AM-GM Inequality

The arithmetic mean–geometric mean inequality (AM-GM) compares the average of nonnegative numbers with the geometric mean of those numbers. It is useful in optimization and in proofs where sums and products appear together.

For $a_1,\ldots,a_n\geq 0$, the inequality is

$$\frac{a_1+\cdots+a_n}{n}\geq (a_1a_2\cdots a_n)^{1/n}.$$

Equality holds if and only if $a_1=\cdots=a_n$. Here are several ways to prove it, from a one-line argument for two numbers to general proofs for any finite number of terms.

Proof for two numbers

For $a,b\geq 0$, a square is always nonnegative:

$$(\sqrt a-\sqrt b)^2\geq 0.$$

Expanding and rearranging gives

$$a+b\geq 2\sqrt{ab}, \qquad\text{so}\qquad \frac{a+b}{2}\geq\sqrt{ab}.$$

The square is zero exactly when $a=b$, which gives the equality condition. This proof is often the quickest way to establish AM-GM for two terms.

Proof by maximizing the product at a fixed sum

Fix a sum $S>0$ and consider all nonnegative $n$-tuples with that sum. This set is closed and bounded, so the product $a_1a_2\cdots a_n$ has a maximum on it. The maximum is positive: the choice $a_i=S/n$ has positive product. Therefore a maximizing tuple cannot have a zero entry.

Suppose two entries of a maximizing tuple are unequal, say $a_i\ne a_j$. Keep their sum fixed and replace both by their average. Their product changes from $a_i a_j$ to

$$\left(\frac{a_i+a_j}{2}\right)^2,$$

which is larger because

$$\left(\frac{a_i+a_j}{2}\right)^2-a_i a_j =\frac{(a_i-a_j)^2}{4}>0.$$

The other entries and the total sum do not change, so this replacement increases the full product. That contradicts maximality. Thus all entries at the maximum must be equal, and the maximum product is $(S/n)^n$.

For any tuple with sum $S$, its product is therefore at most $(S/n)^n$. Taking nonnegative $n$th roots gives

$$(a_1a_2\cdots a_n)^{1/n}\leq \frac{S}{n},$$

which is AM-GM. If $S=0$, every entry is zero and the result is immediate.

Proof using the logarithm

First suppose all $a_i>0$. The natural logarithm is concave on the positive real numbers because

$$(\ln x)''=-\frac{1}{x^2}<0.$$

The concavity inequality for an average says that the logarithm of the arithmetic mean is at least the mean of the logarithms:

$$\ln\left(\frac{a_1+\cdots+a_n}{n}\right) \geq \frac{\ln a_1+\cdots+\ln a_n}{n}.$$

The right-hand side equals $\ln((a_1\cdots a_n)^{1/n})$. Since $\ln$ is increasing, exponentiating proves AM-GM. Strict concavity shows equality is possible only when all the $a_i$ are equal.

If some $a_i=0$, the geometric mean is zero, while the arithmetic mean is nonnegative, so the inequality holds directly. This handles the cases where the logarithm proof cannot be applied.

Worked example

For $2,8,$ and $8$, the arithmetic mean is

$$\frac{2+8+8}{3}=6,$$

while the geometric mean is

$$(2\cdot 8\cdot 8)^{1/3}=128^{1/3}\approx 5.04.$$

Thus $6\geq 5.04$, as AM-GM predicts. The inequality is strict because the three numbers are not all equal.

Common mistakes

• Mixing up the two means: The arithmetic mean is a sum divided by the number of terms; the geometric mean is the corresponding product raised to the reciprocal of that number.
• Forgetting the domain: The standard real-valued statement assumes nonnegative inputs. The logarithm proof specifically requires positive inputs, so handle zero entries separately.
• Claiming equality too broadly: Equality requires every input to be the same, not merely that some pair of inputs match.
• Using a mismatched number of terms: For three numbers, for example, the geometric mean is $(abc)^{1/3}$, not $(abc)^{1/2}$.

Weighted AM-GM

A useful extension assigns nonnegative weights $w_i$ whose sum is $1$. For positive $a_i$, concavity of $\ln$ gives

$$\ln\left(\sum_{i=1}^n w_i a_i\right) \geq \sum_{i=1}^n w_i\ln a_i,$$

so

$$\sum_{i=1}^n w_i a_i\geq \prod_{i=1}^n a_i^{w_i}.$$

The ordinary inequality is the special case $w_i=1/n$. For positive inputs, equality in the weighted version holds when all inputs with positive weight are equal.