Solution
= Solution
Put $\mu=|b|^2>0$ and $\lambda=a\cdot b$. Expanding the nonnegative squared <Euclidean norm> of $c=\mu a-\lambda b$ gives
$$
0\le|c|^2=\mu^2|a|^2-2\mu\lambda^2+\lambda^2|b|^2
=\mu\big(|a|^2|b|^2-(a\cdot b)^2\big).
$$
Dividing by $\mu$ and taking nonnegative <square roots> proves the <Cauchy-Schwarz inequality>:
$$
\boxed{|a\cdot b|\le|a|\,|b|.}
$$
Equality holds exactly when $c=0$, equivalently when the nonzero <vectors> $a,b$ are <linearly dependent>.