Means & Cauchy-Schwarz Inequality

Two of the most powerful and widely-used inequalities in mathematics are the AM-GM inequality and the Cauchy-Schwarz inequality. They appear everywhere — from olympiad problems to optimization, from probability to physics.

Arithmetic Mean and Geometric Mean

📐Definition — AM and GM

For positive real numbers $a$ and $b$ :

The arithmetic mean (AM) is $\dfrac{a + b}{2}$ .
The geometric mean (GM) is $\sqrt{ab}$ .

⚡Theorem — AM-GM Inequality

For all positive reals $a, b$ :

$\frac{a + b}{2} \geq \sqrt{ab}$

with equality if and only if $a = b$ .

Proof.

$(\sqrt{a} - \sqrt{b})^2 \geq 0$ $a - 2\sqrt{ab} + b \geq 0$ $\frac{a + b}{2} \geq \sqrt{ab} \quad \square$

Geometric Interpretation

Consider a semicircle with diameter $a + b$ . Place the diameter along the $x$ -axis, with the dividing point at distance $a$ from the left endpoint. The altitude from this point to the semicircle has length $\sqrt{ab}$ (the geometric mean), while the radius equals $\dfrac{a+b}{2}$ (the arithmetic mean). Since a radius is always at least as long as any altitude within the semicircle, AM $\geq$ GM.

ℹNote — n-Variable Generalization

For positive reals $a_1, a_2, \ldots, a_n$ :

$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}$

with equality iff all $a_i$ are equal. (Stated without proof.)

Applications of AM-GM

✎Example — Minimizing a Sum

For positive $x$ , find the minimum value of $x + \dfrac{1}{x}$ .

Solution. By AM-GM:

$x + \frac{1}{x} \geq 2\sqrt{x \cdot \frac{1}{x}} = 2\sqrt{1} = 2$

Equality holds when $x = \dfrac{1}{x}$ , i.e., $x = 1$ . The minimum value is $\boxed{2}$ .

✎Example — Maximizing a Product

If $a + b = 10$ with $a, b > 0$ , find the maximum value of $ab$ .

Solution. By AM-GM:

$\sqrt{ab} \leq \frac{a + b}{2} = \frac{10}{2} = 5$

Squaring: $ab \leq 25$ . Equality holds when $a = b = 5$ . The maximum value is $\boxed{25}$ .

Cauchy-Schwarz Inequality

⚡Theorem — Cauchy-Schwarz Inequality

For all real numbers $a_1, a_2, b_1, b_2$ :

$(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$

with equality if and only if $\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2}$ (the vectors $(a_1, a_2)$ and $(b_1, b_2)$ are proportional).

Proof. Start with the non-negative square:

$(a_1 b_2 - a_2 b_1)^2 \geq 0$

Expanding:

$a_1^2 b_2^2 - 2a_1 a_2 b_1 b_2 + a_2^2 b_1^2 \geq 0$

Add $a_1^2 b_1^2 + a_2^2 b_2^2$ to both sides:

$a_1^2 b_2^2 + a_2^2 b_1^2 + a_1^2 b_1^2 + a_2^2 b_2^2 \geq 2a_1 a_2 b_1 b_2 + a_1^2 b_1^2 + a_2^2 b_2^2$

Factor:

$(a_1^2 + a_2^2)(b_1^2 + b_2^2) \geq (a_1 b_1 + a_2 b_2)^2 \quad \square$

ℹNote — n-Variable Form

For real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ :

$\left(\sum_{i=1}^{n} a_i b_i\right)^2 \leq \left(\sum_{i=1}^{n} a_i^2\right)\left(\sum_{i=1}^{n} b_i^2\right)$

Application of Cauchy-Schwarz

✎Example — Constrained Optimization

Given $x^2 + y^2 = 1$ , find the maximum value of $3x + 4y$ .

Solution. By Cauchy-Schwarz:

$(3x + 4y)^2 \leq (3^2 + 4^2)(x^2 + y^2) = 25 \cdot 1 = 25$

Therefore $3x + 4y \leq 5$ . The maximum is $\boxed{5}$ , achieved when $\dfrac{x}{3} = \dfrac{y}{4}$ , i.e., $x = \dfrac{3}{5}$ , $y = \dfrac{4}{5}$ .

Practice

✏Exercise — Apply Cauchy-Schwarz

Given $x^2 + y^2 = 4$ , find the maximum value of $2x + 6y$ .

Hint: Apply Cauchy-Schwarz with vectors $(2, 6)$ and $(x, y)$ .