Fix vectors \(\boldv\) and \(\boldw\text{.}\) For any \(t\in\R\) we have by positivity
\begin{equation*}
0\leq \langle t\boldv-\boldw,t\boldv-\boldw\rangle=\langle\boldv,\boldv\rangle t^2-2\langle\boldv,\boldw\rangle t+\langle\boldw,\boldw\rangle=at^2-2bt+c\text{,}
\end{equation*}
where
\begin{equation}
a=\langle\boldv,\boldv\rangle, \ b=\langle\boldv,\boldw\rangle, \ c=\langle\boldw,\boldw\rangle=\norm{w}^2\text{.}\tag{7.1}
\end{equation}
Since \(at^2-2b\,t+c\geq 0\) for all \(t\in \R\) the quadratic polynomial \(p(t)=at^2-2b\,t+c\) has at most one root. Using the quadratic formula we conclude that we must have \(4b^2-4ac\leq 0\text{,}\) since otherwise \(p(t)\) would have two distinct roots. It follows that
\begin{equation*}
4\langle \boldv, \boldw\rangle^2-4\norm{\boldv}^2\norm{\boldw}^2\leq 0\text{,}
\end{equation*}
or equivalently
\begin{equation*}
\langle\boldv,\boldw\rangle^2\leq \norm{\boldv}^2\norm{\boldw}^2\text{.}
\end{equation*}
Taking square-roots yields the desired inequality.
The same reasoning shows that the Cauchy-Schwarz inequality is an actual equality if and only if
\(p(t)=0\) for some
\(t\) if and only if
\(0=\langle t\boldv-\boldw,t\boldv-\boldw\rangle\) if and only if
\(\boldv=t\boldw\) for some
\(t\) (by positivity).
