The Cauchy–Schwarz Inequality

Statement

For any real numbers \[ a_1,\dots,a_n \quad\text{and}\quad b_1,\dots,b_n, \] the Cauchy–Schwarz inequality states that \[ (a_1^2+\cdots+a_n^2)(b_1^2+\cdots+b_n^2) \;\ge\; (a_1b_1+\cdots+a_nb_n)^2. \] Equality holds exactly when the two vectors \[ (a_1,\dots,a_n),\qquad (b_1,\dots,b_n) \] are multiples of each other.

Proof (Using a Non-negative Quadratic Expression)

The key observation is simple but powerful:

For any real number $t$, the sum of squares \[ \sum_{i=1}^n (a_i t - b_i)^2 \] is always non-negative.

Let us expand this expression carefully: \[ (a_i t - b_i)^2 = a_i^2 t^2 \;-\; 2 a_i b_i t \;+\; b_i^2. \] Summing over all $i$ gives a quadratic polynomial in $t$: \[ f(t) = t^2\sum_{i=1}^n a_i^2 \;-\; 2t\sum_{i=1}^n a_i b_i \;+\; \sum_{i=1}^n b_i^2. \] Since $f(t)$ is a sum of squares, we know that \[ f(t) \ge 0 \qquad \text{for all } t\in\mathbb{R}. \]

Now recall a property of quadratic polynomials from high school algebra:

A quadratic $at^2 + bt + c$ is non-negative for all real $t$ only if its discriminant satisfies \[ \Delta = b^2 - 4ac \le 0. \] In our polynomial, \[ a = \sum a_i^2,\qquad b = -2\sum a_i b_i,\qquad c = \sum b_i^2. \] Let us compute the discriminant: \[ \Delta = (-2\sum a_i b_i)^2 - 4\left(\sum a_i^2\right)\left(\sum b_i^2\right) \;\le\; 0. \] Expanding the first term gives \[ 4\left(\sum a_i b_i\right)^2 - 4\left(\sum a_i^2\right)\left(\sum b_i^2\right) \le 0. \] Dividing by 4 yields \[ \left(\sum a_i b_i\right)^2 \le \left(\sum a_i^2\right)\left(\sum b_i^2\right), \] which is exactly the Cauchy–Schwarz inequality.

When Does Equality Occur?

Equality in the discriminant condition occurs precisely when \[ f(t) = \sum (a_i t - b_i)^2 \] touches zero for some real number $t$.

But a sum of squares equals zero only when each term is zero: \[ a_i t = b_i \qquad\text{for all } i. \] This means that \[ (b_1,\dots,b_n) = t\,(a_1,\dots,a_n), \] so the two vectors are multiples of each other, exactly the equality condition of Cauchy–Schwarz.