Cauchy–Schwarz Inequality

Step 1: Set up an auxiliary quadratic. Consider the real variable $t$ and the expression $\sum_{i=1}^n (a_i t - b_i)^2$. Since it is a sum of squares of real numbers, it is always $\ge 0$ for every real $t$: $\sum(a_i t - b_i)^2 \ge 0$.

**Step 2: Expand into a quadratic in $t$.** Expanding, $\sum (a_i t - b_i)^2 = t^2 \sum a_i^2 - 2t \sum a_i b_i + \sum b_i^2$. Write $A = \sum a_i^2$, $B = \sum a_i b_i$, $C = \sum b_i^2$, so the expression is $At^2 - 2Bt + C \ge 0$ for all real $t$.

Step 3: Handle the degenerate case. If $A = 0$ then every $a_i = 0$, so both sides of the claimed inequality $B^2 \le AC$ are $0$, and the inequality holds trivially (with equality).

**Step 4: Use the discriminant when $A > 0$.** A quadratic $At^2 - 2Bt + C$ with positive leading coefficient $A$ that is $\ge 0$ for every real $t$ can have at most one real root, so its discriminant cannot be positive: $(2B)^2 - 4AC \le 0$, i.e. $4B^2 \le 4AC$, i.e. $B^2 \le AC$.

Step 5: Conclude. Substituting back, $\left(\sum a_i b_i\right)^2 \le \left(\sum a_i^2\right)\left(\sum b_i^2\right)$, which is exactly $\left(\sum a_i^2\right)\left(\sum b_i^2\right) \ge \left(\sum a_i b_i\right)^2$. Equality holds exactly when the discriminant is zero, i.e. the quadratic has a real double root $t_0$, i.e. $a_i t_0 = b_i$ for every $i$ — the sequences $(a_i)$ and $(b_i)$ are proportional.