Understanding a Certain Proof of the Cauchy-Schwarz Inequality Cauchy-Schwarz Proof Outline:


*

*Let $u,v \in V$ s.t. $u,v \ne 0$.

*Then $0 \le \| u - rv\|^2$.

*Let $r = {\overline{\left\langle u,v\right\rangle} \over \left\langle v,v \right\rangle}$.

*Then one obtains that
$$
0 \le \| u \|^2 - {|\left\langle u, v \right\rangle|^2 \over \|v\|^2}
$$
which then implies that
$$
|\left\langle u,v \right\rangle| \le \|u \| \|v\|
$$
as desired.
Question: Where did $r = {\overline{\left\langle u,v \right\rangle} \over \left\langle v, v \right\rangle}$ come from?  What does that represent?  It seems closely related -- but not equal to -- the notion of $u$'s projection onto $v$ in the sense that $r$'s conjugate would represent this scalar.
 A: This is orthogonal projection, assuming $v \ne 0$. This is because
$$
          \left[u - \left(u,\frac{1}{\|v\|}v\right)\frac{1}{\|v\|}v\right] \perp v.
$$
An equivalent way to write this projection is as
$$
     u - \left(u,\frac{1}{\|v\|}v\right)\frac{1}{\|v\|}v=u-\frac{(u,v)}{(v,v)}v.
$$
The orthogonal projection of $u$ onto the "line" through the origin in the direction of $v$ is the point $\frac{(u,v)}{(v,v)}v$ on that line. The distance from $u$ to the line $\{ \lambda v : \lambda \mbox{ is a scalar}\}$ is minimized when $\lambda=(u,v)/(v,v)$. By the "Pythagorean Theorem",
$$
\begin{align}
 \|u\|^{2}=\left\|\left(u-\frac{(u,v)}{(v,v)}v\right)+\frac{(u,v)}{(v,v)}v\right\|^{2}
   & = \left\|u-\frac{(u,v)}{(v,v)}v\right\|^{2}+\frac{|(u,v)|^{2}}{|(v,v)|^{2}}\|v\|^{2} \\
   & \ge \frac{|(u,v)|^{2}}{|(v,v)|^{2}}\|v\|^{2}=\frac{|(u,v)|^{2}}{\|v\|^{2}}.
\end{align}
$$
That gives you the Cauchy-Schwarz inequality if $v \ne 0$, and you can see that you have equality between the first and last terms iff $u-\frac{(u,v)}{(v,v)}v=0$.
