Math Core

Many AIME problems ask for "the minimum value" or "the largest possible value." Calculus is rarely the fastest route. Instead, a classical inequality gives a bound, and then you show the bound is actually reached. Both halves matter: a bound you can't achieve is not an answer.

AM-GM

The AM-GM inequality

For nonnegative reals $x_1, \dots, x_n$ ,

\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n},

with equality exactly when $x_1 = x_2 = \dots = x_n$ .

The two-variable case is just a square being nonnegative: $x + y - 2\sqrt{xy} = (\sqrt x - \sqrt y)^2 \ge 0$ .

AM-GM is useful in two directions.

Fixed product, minimize a sum. If $xy$ is constant, $x + y \ge 2\sqrt{xy}$ is a constant lower bound.
Fixed sum, maximize a product. If $x + y$ is constant, $xy \le \left(\dfrac{x+y}{2}\right)^2$ .

The skill is splitting terms so the equality case can happen. To maximize $x^2y^3$ given $x + y = 5$ , don't apply AM-GM to $x$ and $y$ . Split $x$ into two equal halves and $y$ into three equal thirds, so that the five pieces $\dfrac x2, \dfrac x2, \dfrac y3, \dfrac y3, \dfrac y3$ still add to a constant and can all be equal.

Cauchy-Schwarz

For reals $a_i, b_i$ ,

(a_1b_1 + a_2b_2 + \dots + a_nb_n)^2 \le (a_1^2 + \dots + a_n^2)(b_1^2 + \dots + b_n^2),

with equality when the vectors $(a_i)$ and $(b_i)$ are proportional. Geometrically it says a dot product is at most the product of the lengths. A very handy form for fractions (sometimes called Engel's form or Titu's lemma) is

\frac{u_1^2}{v_1} + \frac{u_2^2}{v_2} + \dots + \frac{u_n^2}{v_n} \ge \frac{(u_1 + u_2 + \dots + u_n)^2}{v_1 + v_2 + \dots + v_n} \qquad (v_i > 0).

Other tools

Completing the square. A quadratic in several variables can be written as a sum of squares plus a constant. The constant is the minimum.
The discriminant. If $a + b = s$ and $ab = p$ are real, then $a, b$ are roots of $t^2 - st + p$ , so $s^2 \ge 4p$ . This turns "how large can $c$ be?" into a quadratic inequality in $c$ .
Geometry. $\sqrt{x^2 + a^2}$ is the length of a segment. A sum of such lengths is a broken path, and the shortest path is a straight line.

Common mistake

Always check the equality case. For example, AM-GM gives $x + \dfrac{1}{x} \ge 2$ , but if the problem requires $x \ge 3$ , equality ( $x = 1$ ) is impossible, and the true minimum is $3 + \dfrac13$ . If your equality conditions contradict each other or the constraints, the bound is not attained.

Worked examples

Worked example: Fixed product

Find the minimum of $x + \dfrac{36}{x}$ for $x > 0$ .

By AM-GM, $x + \dfrac{36}{x} \ge 2\sqrt{36} = 12$ , with equality when $x = \dfrac{36}{x}$ , that is, $x = 6$ . The minimum is $12$ .

Worked example: Splitting terms

Positive reals $x, y$ satisfy $x + y = 3$ . Find the maximum of $xy^2$ .

Split $y$ in half: $x + \dfrac y2 + \dfrac y2 = 3$ . By AM-GM on these three terms,

1 = \frac{x + \frac y2 + \frac y2}{3} \ge \sqrt[3]{\frac{xy^2}{4}},

so $xy^2 \le 4$ . Equality needs $x = \dfrac y2$ , which with $x + y = 3$ gives $x = 1$ , $y = 2$ , and indeed $1 \cdot 4 = 4$ . The maximum is $4$ .

Worked example: Cauchy-Schwarz

Real numbers $x, y$ satisfy $x^2 + y^2 = 25$ . Find the maximum of $3x + 4y$ .

By Cauchy-Schwarz, $(3x + 4y)^2 \le (3^2 + 4^2)(x^2 + y^2) = 25 \cdot 25$ , so $3x + 4y \le 25$ . Equality needs $(x, y)$ proportional to $(3, 4)$ : $(x, y) = (3, 4)$ gives $9 + 16 = 25$ . The maximum is $25$ .

Worked example: How large can one variable be?

Real numbers $a, b, c$ satisfy $a + b + c = 6$ and $ab + bc + ca = 9$ . Find the largest possible value of $c$ .

Then $a + b = 6 - c$ and $ab = 9 - c(a + b) = 9 - c(6 - c)$ . Since $a, b$ are real, $(a + b)^2 \ge 4ab$ :

(6 - c)^2 \ge 4(9 - 6c + c^2) \iff 36 - 12c + c^2 \ge 36 - 24c + 4c^2 \iff 3c^2 - 12c \le 0.

So $0 \le c \le 4$ . At $c = 4$ : $a + b = 2$ and $ab = 1$ , so $a = b = 1$ , which works. The maximum is $4$ .