Arithmetic and Geometric Means; AM-GM inequality

Of the two numbers $4$ and $9$, the arithmetic mean is $6.5$ and the geometric mean is $6$. Notice that AM is strictly larger than GM. That observation, generalised, becomes one of the most powerful inequalities in elementary mathematics , the arithmetic-geometric mean inequality, or AM-GM for short.

Definitions

The arithmetic mean (AM) of two numbers $a, b$ is $A = \frac{a + b}{2}.$

The geometric mean (GM) of two positive numbers $a, b$ is $G = \sqrt{ab}.$

The AM is the midpoint between $a$ and $b$ on the number line. The GM is the "midpoint" in a multiplicative sense , the number whose ratio to $a$ equals its ratio to $b$.

More generally, the AM of $n$ numbers is $\dfrac{a_1 + a_2 + \dots + a_n}{n}$ and the GM (when all are positive) is $(a_1 a_2 \cdots a_n)^{1/n}$.

Inserting means between two numbers

$n$ arithmetic means between $a$ and $b$: numbers $A_1, A_2, \dots, A_n$ such that $a, A_1, \dots, A_n, b$ form an AP. Then the common difference is $d = \dfrac{b - a}{n + 1}$ and $A_k = a + k d$.

$n$ geometric means between $a$ and $b$ (both positive): numbers $G_1, \dots, G_n$ such that $a, G_1, \dots, G_n, b$ form a GP. Then $r = (b/a)^{1/(n+1)}$ and $G_k = a r^k$.

The AM-GM inequality

Theorem. For positive real numbers $a, b$: $\boxed{\frac{a + b}{2} \ge \sqrt{ab},}$ with equality if and only if $a = b$.

Proof. $\left(\sqrt{a} - \sqrt{b}\right)^2 \ge 0$. Expand: $a - 2\sqrt{ab} + b \ge 0$, i.e. $a + b \ge 2\sqrt{ab}$, hence $\dfrac{a+b}{2} \ge \sqrt{ab}$. Equality iff $\sqrt{a} = \sqrt{b}$, i.e. $a = b$. $\qed$

For $n$ positive numbers, $\frac{a_1 + a_2 + \dots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \cdots a_n}.$

Why AM-GM matters

It turns sum problems into product problems and vice versa. Whenever you want to maximise a product given a fixed sum (or minimise a sum given a fixed product), AM-GM gives the answer instantly , the extremum occurs when all variables are equal.

Standard application. What is the minimum value of $x + \dfrac{1}{x}$ for $x > 0$?

By AM-GM: $x + \dfrac{1}{x} \ge 2 \sqrt{x \cdot \dfrac{1}{x}} = 2$, with equality at $x = 1$.

Worked examples

Example 1. Insert three arithmetic means between $2$ and $14$.

$n = 3$, so $d = \dfrac{14 - 2}{4} = 3$. Means: $5, 8, 11$.

Example 2. Insert three geometric means between $2$ and $32$.

$r = (32/2)^{1/4} = 16^{1/4} = 2$. Means: $4, 8, 16$.

Example 3. If $a, b, c$ are in AP and $a, b, c$ are positive, show $b \ge \sqrt{ac}$.

$a, b, c$ in AP means $b = \dfrac{a + c}{2}$. By AM-GM, $\dfrac{a + c}{2} \ge \sqrt{ac}$. So $b \ge \sqrt{ac}$, with equality iff $a = c$.

Example 4. Find the minimum value of $\dfrac{a}{b} + \dfrac{b}{a}$ for $a, b > 0$.

By AM-GM: $\dfrac{a/b + b/a}{2} \ge \sqrt{\dfrac{a}{b} \cdot \dfrac{b}{a}} = 1$. So the minimum is $2$, attained at $a = b$.

Example 5. If positive numbers $a, b, c$ have $a + b + c = 12$, find the maximum value of $abc$.

By AM-GM, $\dfrac{a + b + c}{3} \ge \sqrt[3]{abc}$, so $4 \ge \sqrt[3]{abc}$, i.e. $abc \le 64$. Maximum is $64$, at $a = b = c = 4$.

Try it yourself

1. Find the arithmetic mean of $11$ and $29$.
2. Find the geometric mean of $4$ and $25$.
3. Insert two arithmetic means between $3$ and $24$.
4. Insert four geometric means between $\dfrac{1}{2}$ and $16$.
5. Show that if $x > 0$, then $x + \dfrac{4}{x} \ge 4$.
6. Find the minimum of $\dfrac{1}{a} + \dfrac{1}{b}$ given $a + b = 4$, $a, b > 0$. (Answer: $1$, at $a = b = 2$.)
7. If $A$ and $G$ are AM and GM of two positive numbers, prove $A \ge G$.
8. If $A$ and $G$ are AM and GM of two positive numbers $a, b$, show that the numbers are $A \pm \sqrt{A^2 - G^2}$.
9. Prove that for $a, b, c > 0$, $(a + b)(b + c)(c + a) \ge 8 abc$.
10. Find the maximum value of $xy$ given $x + y = 10$, $x, y > 0$.
11. Find the minimum value of $\dfrac{(1 + x)(1 + y)}{xy}$ for $x, y > 0$ given $x + y = 1$.
12. Show $a^2 + b^2 + c^2 \ge ab + bc + ca$ for all real $a, b, c$.

Pitfalls / Tricks

• AM-GM requires the numbers to be positive (or non-negative).
• Equality in AM-GM happens only when all numbers are equal , this often pinpoints the location of an extremum.
• For optimisation, set the variables to make the AM-GM bound tight; that point is the extremum.
• Insight. AM-GM is the calculus-free way to find extrema. Many JEE inequality problems collapse to a single application of it.