# Arithmetic Mean-Geometric Mean Inequality

The Arithmetic Mean-Geometric Mean Inequality (AM-GM or AMGM) is an elementary inequality, generally one of the first ones taught in inequality courses.

## Theorem

The AM-GM states that for any set of positive real numbers, the arithmetic mean of the set is greater than or equal to the geometric mean of the set. Algebraically, this is written:

For a set of nonnegative real numbers $a_1,a_2,\ldots,a_n$, the following always holds:

$\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)\geq\sqrt[n]{a_1a_2\cdots a_n}$

Using the shorthand notation for summations and products:

$\left(\frac{\sum\limits_{i=1}^{n}a_i}{n}\right)\geq\sqrt[n]{\prod\limits_{i=1}^{n}a_i}$

For example, for the set $\{9,12,54\}$, the Arithmetic Mean, 25, is greater than the Geometric Mean, 18; AM-GM guarantees this is always the case.

The equality condition of this inequality states that the arithmetic mean and geometric mean are equal if and only if all members of the set are equal.

AMGM can be used fairly frequently to solve Olympiad-level inequality problems, such as those on the USAMO and IMO.

### Weighted Form

The weighted form of AMGM is given by using weighted averages. For example, the weighted arithmetic mean of $x$ and $y$ with $3:1$ is $\frac{3y+x}{2\cdot 3}$ and the geometric is $\sqrt[2+3]{xy^3}$. AMGM applies to weighted averages.

## Problems

### Intermediate

• Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$.

(Source)

• Let $a$, $b$, and $c$ be positive real numbers. Prove that
$(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3$.