Intuition on proof of Cauchy Schwarz inequality

Pick $k$ so as to minimize the distance from $kx$ to $y$, or rather its square, noting that $$\def\Re{\operatorname{Re}} 0\le(y-kx,y-kx)=(x,x)k^2-2\Re((x,y)k)+(y,y)$$ for all $k$. Now pick the $k$ that minimizes the right hand side (in order to get the most out of the inequality), and find that it is the very same $k$ used in the standard proof.

You can determine this $k$ by standard calculus methods, or by completing the square. But first, pick the phase of $k$ to make $(x,y)k$ positive, thus avoiding the difficulty of working with the real part.

This is similar to Harald's answer but there is no explicit choice of $k$. For simplicity let us take the real case. Fix vectors $x$ and $y$, and consider the quadratic polynomial $$ p(k)=(x+ky,x+ky)=(x,x)+2k(x,y)+k^2(y,y)\geq0. $$ The polynomial must be nonnegative for any value of $k\in\mathbb{R}$, meaning that the equation $p(k)=0$ has at most one solution. This can be expressed in terms of the discriminant as $$ D = (x,y)^2-(x,x)(y,y)\leq0, $$ giving the Cauchy-Bunyakowsky-Scwarz inequality.

The moral of the story is that the choice of $k$ is mainly due to the fact that there is a quadratic polynomial behind the proof.