The intuition behind the Schwarz inequality is perhaps easily seen for an inner product over a real vector space; for any vectors $u, v$,
\begin{align}
\langle u \pm v, u \pm v\rangle \geqslant 0.
\end{align}
Now take $u, v$ to have unit length and expand the product to obtain,
\begin{align}
2 \geqslant \pm 2 \langle u, v \rangle
\end{align}
so that $\lvert \langle u,v \rangle \rvert \leqslant 1$. This is extended to general vectors $u$ and $v$ by dividing each by its norm and using the bi-linearity of the inner product.
The argument is also extended to a complex linear space by taking again unit vectors $u$ and $v$ but now multiply $u$ by $e^{i\theta}$ where $\theta$ is chosen to make $\langle u,v \rangle$ real and positive. This is always possible.
Then
\begin{align}
\langle e^{i\theta}u - v, e^{i\theta} u - v \rangle &= \langle e^{i\theta} u, e^{i \theta}u\rangle - e^{i\theta} \langle u, v\rangle
- e^{-i\theta}\langle v,u\rangle +\langle v,v \rangle \\
&=\langle u,u \rangle - 2 \Re \big( e^{i\theta}\langle u,v \rangle\big) + \langle v,v \rangle
\end{align}
Now the choice of $\theta$ comes into play. We have made $e^{i\theta}\langle u,v\rangle$ real and positive, so as before we now obtain
\begin{align}
\Re \big( e^{i\theta} \langle u,v \rangle \big) &= e^{i\theta} \langle u,v \rangle \\
&= \lvert e^{i\theta} \langle u,v \rangle \rvert \\
&= \lvert \langle u,v \rangle \rvert
\end{align}
leading neatly to $\lvert \langle u,v \rangle \rvert \leqslant 1$. This is extended to general $u$ and $v$ as before.
