= Solution
Write $A=\sum_i a_i^2$, $B=\sum_i b_i^2$, $C=\sum_i a_ib_i$. If $B=0$, every $b_i=0$ and the claimed inequality is immediate. Otherwise the sum of squares
$$
0\leq\sum_i(a_i-tb_i)^2=A-2tC+t^2B
$$
holds for every real $t$. Choosing $t=C/B$ gives $0\leq A-C^2/B$, and hence the <Cauchy-Schwarz inequality> in its stronger absolute-value form:
$$
\boxed{|C|\leq\sqrt{AB}.}
$$
In particular $C\leq\sqrt A\sqrt B$, as requested. For unit <vectors>, $A=B=1$, so this same argument gives
$$
\boxed{|a\cdot b|\leq1.}
$$
The absolute value is essential for opposite-pointing <vectors>; it follows from the squared inequality without a sign assumption on their <dot product>.
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