Cauchy-Schwarz Inequality - How can you divine the linchpin of $\Vert proj_{W}\mathbf{v}\Vert$? Source: Linear Algebra by Lay (4 edn 2011). p. 379. Section 6.7. Theorem 16.


If $\mathbf{u}=\mathbf{0}$, then the inequality becomes $ 0 \le 0 $, which is true. See Practice Problem 6.7.1 on p. 382. 
If $\mathbf{u}\neq \mathbf{0}$, let $W$ be the subspace spanned by $\mathbf{u}$. 

$1.$ How can you divine to define $W$ like this? I'm not clairvoyant enough to spot this  flash of genius.

$$
\Vert proj_{W}\mathbf{v}\Vert=\Vert\frac{\langle v,u\rangle}{\langle u,u\rangle}u\Vert=\frac{|\langle v,u\rangle|}{|\langle u,u\rangle|}\Vert u\Vert=\frac{|\langle v,u\rangle|}{||u||^{2}}\Vert u\Vert=\frac{|\langle u,v\rangle|}{\Vert u\Vert}
$$
  Since $\Vert proj_{W}\mathbf{v}\Vert^2 \leq\Vert v\Vert^2 \implies \Vert proj_{W}v\Vert\leq\Vert v\Vert$, so $\displaystyle \frac{|\langle u,v\rangle|}{\Vert u\Vert}\leq\Vert v\Vert$.

$2.$ How can you divine this proof's linchpin: computing $\Vert proj_{W}v\Vert$? 
 A: This is just the geometric interpretation of the usual proof. The line in direction $\mathbf u$ is parametrized as $$W=\{t \mathbf u:t\in\Bbb R\},$$ the projection $P_W(\mathbf v)$ is the point of minimal distance $\|\mathbf v-t\mathbf u\|$ from $\mathbf v$ to a point on the line $W$, so
$$
P_W(\mathbf v)
=\arg\min\{ \|\mathbf w-\mathbf v\|:\mathbf w\in W\}
=\arg\min\{\|t\mathbf u-\mathbf v\|:t\in\Bbb R\}\cdot \mathbf u,
$$
 this distance becomes minimal if the square of it is minimal, the square is a quadratic polynomial in $t$
$$
\|\mathbf v-t\mathbf u\|^2
=\|\mathbf v\|^2-2t\,\langle \mathbf u,\mathbf v\rangle +t^2\,\|\mathbf u\|^2
$$
and the minimal point of this parabola can be determined in different ways.
For instance by completing the square one obtains
$$
0\le \|\mathbf u\|^2\,\|\mathbf v-t\mathbf u\|^2
= \|\mathbf u\|^2\,\|\mathbf v\|^2-\langle \mathbf u,\mathbf v\rangle^2+(t\|\mathbf u\|^2-\langle \mathbf u,\mathbf v\rangle)^2
$$
which gives the minimizer $t$ and the Cauchy-Schwarz inequality at the same time.

Specifically to the questions:


*

*Why this line and the projection: This is just the way this proof works. It has the additional advantage that it is easily extendable to general Hilbert spaces. One could also do a non-geometrical, purely analytical proof in the fashion the general Hölder inequality, of which Cauchy-Schwarz is a special case, is proved.

*What is projection: In the case of normed spaces the point of least distance, as used above. There is also central projection as in photography which is only loosely related to this kind of projection.

Another answer to 2): Assume $u\ne 0$. Then the space can be divided up into a family of planes $$E_d=\{\mathbf v:⟨\mathbf u,\mathbf v⟩=d\}.$$ The point on the line $W$ through $\mathbf u$ that is on the plane is $\mathbf u_d=\frac{d}{\|\mathbf u\|^2}\mathbf u$. By construction, $\mathbf u$ is the normal of the plane $E_d$, i.e., the plane is perpendicular to the line $W$.
Thus under orthogonal projection all points in $E_d$ get mapped to $\mathbf u_d$. Eliminating $d$ from this relation results in
$$
\operatorname{proj}_W \mathbf v =\frac{⟨\mathbf u,\mathbf v⟩}{\|\mathbf u\|^2}\mathbf u
$$
