Math Core

"What is the smallest possible value of…" is one of the most common question stems on the AMC 10 and 12. Calculus isn't expected. Instead, the go-to tool is the AM-GM inequality, together with the simple fact that squares are never negative. The skill is in arranging an expression so that the inequality fits, and in checking that equality can actually happen.

Squares are nonnegative

Every inequality in this module comes from one fact: $t^2 \ge 0$ for every real $t$ , with equality only when $t = 0$ . For example, completing the square shows

x^2 - 6x + 13 = (x - 3)^2 + 4 \ge 4,

with equality at $x = 3$ . So the minimum value is $4$ .

The AM-GM inequality

Expand $(\sqrt{a} - \sqrt{b})^2 \ge 0$ for $a, b \ge 0$ : you get $a + b \ge 2\sqrt{ab}$ . The same idea extends to any number of terms.

AM-GM

For nonnegative numbers $a_1, a_2, \dots, a_n$ ,

\frac{a_1 + a_2 + \dots + a_n}{n} \ \ge\ \sqrt[n]{a_1 a_2 \cdots a_n},

with equality exactly when all the numbers are equal.

If the product is fixed, the sum is smallest when the terms are equal.
If the sum is fixed, the product is largest when the terms are equal.

Worked example: A classic minimum

Find the minimum of $x + \dfrac{9}{x}$ for $x > 0$ .

The product $x \cdot \dfrac{9}{x} = 9$ is constant, so AM-GM gives $x + \dfrac{9}{x} \ge 2\sqrt{9} = 6$ . Equality needs $x = \dfrac{9}{x}$ , so $x = 3$ , which is allowed. The minimum is $6$ .

y = x + 9/x touches its minimum value 6 at x = 3.Open in grapher →

Making the product constant

AM-GM only gives a useful bound when the product of the terms (or the sum, for a maximum) is a constant. If it isn't, rearrange: scale a term, or split a term into equal pieces.

Worked example: Scaling to match a constraint

Positive numbers satisfy $2x + y = 12$ . What is the largest possible value of $xy$ ?

The fixed sum is $2x + y$ , so apply AM-GM to $2x$ and $y$ : $\sqrt{2x \cdot y} \le \dfrac{2x + y}{2} = 6$ , so $2xy \le 36$ and $xy \le 18$ . Equality when $2x = y = 6$ , so $x = 3$ , $y = 6$ .

Worked example: Splitting a term

Find the minimum of $x^2 + \dfrac{16}{x}$ for $x > 0$ .

The product $x^2 \cdot \dfrac{16}{x}$ is not constant. Split $\dfrac{16}{x}$ into two equal halves so the $x$ 's cancel:

x^2 + \frac{8}{x} + \frac{8}{x} \ \ge\ 3\sqrt[3]{x^2 \cdot \frac{8}{x} \cdot \frac{8}{x}} = 3\sqrt[3]{64} = 12.

Equality needs $x^2 = \dfrac{8}{x}$ , so $x = 2$ . Check: $4 + 8 = 12$ . The minimum is $12$ .

Multiplying by 1 in disguise

When you have a constraint like $\dfrac{1}{x} + \dfrac{4}{y} = 1$ and want to minimize $x + y$ , multiply by the constraint (which equals $1$ ):

Worked example: Using the constraint

Positive numbers satisfy $\dfrac{1}{x} + \dfrac{4}{y} = 1$ . Find the minimum of $x + y$ .

x + y = (x + y)\left(\frac{1}{x} + \frac{4}{y}\right) = 1 + 4 + \frac{y}{x} + \frac{4x}{y} \ \ge\ 5 + 2\sqrt{4} = 9.

Equality when $\dfrac{y}{x} = \dfrac{4x}{y}$ , so $y = 2x$ . Then $\dfrac{1}{x} + \dfrac{2}{x} = 1$ gives $x = 3$ , $y = 6$ , and $x + y = 9$ .

Common mistake

An AM-GM bound is only the answer if equality can happen. For instance, $x + \frac{1}{x} \ge 2$ for $x > 0$ , but on the interval $x \ge 3$ equality ( $x = 1$ ) is impossible, and the true minimum is $3 + \frac{1}{3}$ . Always find the equality case and check it satisfies every condition in the problem.

Tip

A related tool is the Cauchy–Schwarz inequality: $(a^2 + b^2)(c^2 + d^2) \ge (ac + bd)^2$ , with equality when $\frac{a}{c} = \frac{b}{d}$ . It's the quickest way to minimize $x^2 + y^2$ given a linear condition on $x$ and $y$ .

Practice

Practice 1

What is the minimum value of $4x + \dfrac{25}{x}$ for $x > 0$ ?

Practice 2

Positive numbers satisfy $x + 3y = 12$ . What is the largest possible value of $xy$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Real numbers satisfy $x + 2y = 10$ . What is the smallest possible value of $x^2 + y^2$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Positive numbers $a, b, c$ satisfy $abc = 8$ . What is the smallest possible value of $(1 + a)(1 + b)(1 + c)$ ?

Practice 5

What is the minimum value of $x^3 + \dfrac{48}{x}$ for $x > 0$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Positive numbers satisfy $\dfrac{1}{x} + \dfrac{9}{y} = 1$ . What is the smallest possible value of $x + y$ ?

Practice 7

An open-top rectangular box has a square base and a volume of $32$ cubic units. What is the smallest possible total area of its base and four sides?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Positive numbers satisfy $x + y + z = 6$ . What is the largest possible value of $x y^2 z^3$ ?

Real contest practice

2000 AMC 12, Problem 12: a product with fixed sum is largest when the factors are equal.
2016 AMC 12B, Problem 14: minimize the sum of a geometric series, a one-variable optimization.