 Solution

Write $A={\sum }_i{a}_i^2$, $B={\sum }_i{b}_i^2$, $C={\sum }_i{a}_i{b}_i$. If $B=0$, every $b_i=0$ and the claimed inequality is immediate. Otherwise the sum of squaresholds for every real $t$. Choosing $t=C/B$ gives $0\le A-C^2/B$, and hence the Cauchy-Schwarz inequality in its stronger absolute-value form:In particular $C\le \sqrt{A}\sqrt{B}$, as requested. For unit vectors, $A=B=1$, so this same argument givesThe absolute value is essential for opposite-pointing vectors; it follows from the squared inequality without a sign assumption on their dot product.