He does it because it works. Essentially, as you see, $$\sum_{j=1}^n |Ba_j-Cb_j|^{2}$$ is always greater or equal to zero. He then shows that $$\tag 1 \sum_{j=1}^n |Ba_j-Cb_j|^{2}=B(AB-|C|^2)$$
and having assumed $B>0$; this means $AB-|C|^2\geq 0$, which is the Cauchy Schwarz inequality.
ADD Let's compare two different proofs of Cauchy Schwarz in $\Bbb R^n$.
PROOF1. We can see the Cauchy Schwarz inequality is true whenever ${\bf x}=0 $ or ${\bf{y}}=0$, so discard those. Let ${\bf x}=(x_1,\dots,x_n)$ and ${\bf y }=(y_1,\dots,y_n)$, so that $${\bf x}\cdot {\bf y}=\sum_{i=1}^n x_iy_i$$
We wish to show that $$|{\bf x}\cdot {\bf y}|\leq ||{\bf x}||\cdot ||{\bf y}||$$
Define $$X_i=\frac{x_i}{||{\bf x}||}$$
$$Y_i=\frac{y_i}{||{\bf y}||}$$
Because for any $x,y$ $$(x-y)^2\geq 0$$ we have that $$x^2+y^2\geq 2xy$$ Using this with $X_i,Y_i$ for $i=1,\dots,n$ we have that $$X_i^2 + Y_i^2 \geqslant 2{X_i}{Y_i}$$
and summing up through $1,\dots,n$ gives $$\eqalign{
& \frac{{\sum\limits_{i = 1}^n {y_i^2} }}{{||{\bf{y}}|{|^2}}} + \frac{{\sum\limits_{i = 1}^n {x_i^2} }}{{||{\bf{x}}|{|^2}}} \geqslant 2\frac{{\sum\limits_{i = 1}^n {{x_i}{y_i}} }}{{||{\bf{x}}|| \cdot ||{\bf{y}}||}} \cr
& \frac{{||{\bf{y}}|{|^2}}}{{||{\bf{y}}|{|^2}}} + \frac{{||{\bf{x}}|{|^2}}}{{||{\bf{x}}|{|^2}}} \geqslant 2\frac{{\sum\limits_{i = 1}^n {{x_i}{y_i}} }}{{||{\bf{x}}|| \cdot ||{\bf{y}}||}} \cr
& 2 \geqslant 2\frac{{\sum\limits_{i = 1}^n {{x_i}{y_i}} }}{{||{\bf{x}}|| \cdot ||{\bf{y}}||}} \cr
& ||{\bf{x}}|| \cdot ||{\bf{y}}|| \geqslant \sum\limits_{i = 1}^n {{x_i}{y_i}} \cr} $$
NOTE How may we add the absolute value signs to conclude?
PROOF2
We can see the Cauchy Schwarz inequality is true whenever ${\bf x}=0 $ or ${\bf{y}}=0$, or $y=\lambda x$ for some scalar. Thus, discard those hypotheses. Then consider the polynomial (here $\cdot$ is inner product) $$\displaylines{
P(\lambda ) = \left\| {{\bf x} - \lambda {\bf{y}}} \right\|^2 \cr
= ( {\bf x} - \lambda {\bf{y}})\cdot({\bf x} - \lambda {\bf{y}}) \cr
= {\left\| {\bf x} \right\|^2} - 2\lambda {\bf x} \cdot {\bf{y}} + {\lambda ^2}{\left\| {\bf{y}} \right\|^2} \cr} $$
Since ${\bf x}\neq \lambda{\bf y}$ for any $\lambda \in \Bbb R$, $P(\lambda)>0$ for each $\lambda\in\Bbb R$. It follows the discriminant is negative, that is $$\Delta = b^2-4ac={\left( {-2\left( {{\bf x} \cdot y} \right)} \right)^2} - 4{\left\| {\bf x} \right\|^2}{\left\| {\bf{y}} \right\|^2} <0$$ so that $$\displaylines{
{\left( {{\bf x}\cdot {\bf{y}}} \right)^2} <{\left\| {\bf x} \right\|^2}{\left\| {\bf{y}} \right\|^2} \cr
\left| {{\bf x} \cdot {\bf{y}}} \right| <\left\| {\bf x}\right\| \cdot \left\| {\bf{y}} \right\| \cr} $$ which is Cauchy Schwarz, with equaliy if and only if ${\bf x}=\lambda {\bf y}$ for some $0\neq \lambda \in\Bbb R$ or either vector is null.
One proof shows the Cauchy Schwarz inequality is a direct consequence of the known fact that $x^2\geq 0$ for each real $x$. The other is shorter and sweeter, and uses the fact that a norm is always nonnegative, and properties of the inner product of vectors in $\Bbb R^n$, plus that fact that a polynomial in $\Bbb R$ with no real roots must have negative discriminant.
Best Answer
Well, since no one gave a complete answer yet--and because I wrote one anyway--here's the proof by induction, in a manner which is hopefully easy for students (without much proof experience) to understand. Credit goes to the Wu and Wu paper posted by @Jeff.
Both sides of the Schwarz inequality are real numbers $\geq 0$. If $\sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2 = 0$, then it must be that $a_1 = a_2 = \ldots = a_n = 0$ and/or $b_1 = b_2 = \ldots = b_n = 0$, so clearly $|\sum_{j=1}^n a_j \overline{b_j}|^2$ also $= 0$ and we are done. Now we only need to prove the case in which both sides of the inequality are positive.
Base Case. For $n = 1$, we have $$|\sum_{j=1}^1 a_j \overline{b_j}|^2 = |a_j \overline{b_j}|^2 = |a_j|^2 |b_j|^2 = \sum_{j=1}^1 |a_j|^2 \sum_{j=1}^1 |b_j|^2.$$
Inductive Step. The inductive hypothesis is $|\sum_{j=1}^{n-1} a_j \overline{b_j}|^2 \leq \sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2$. Since we only need to worry about the case in which both sides are positive, so we can take the square root to obtain $$|\sum_{j=1}^{n-1} a_j \overline{b_j}| \leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2}.$$
Thus $|\sum_{j=1}^n a_j \overline{b_j}|$
$= |\sum_{j=1}^{n-1} a_j \overline{b_j} + a_n \overline{b_n}|$
$\leq |\sum_{j=1}^{n-1} a_j \overline{b_j}| + |a_n \overline{b_n}|$ (by the triangle inequality)
$\leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2} + |a_n \overline{b_n}|$ (by the inductive hypothesis)
$= \sqrt{\sum_{j=1}^{n-1} |a_j|^2} \sqrt{\sum_{j=1}^{n-1} |b_j|^2} + |a_n| |b_n|.$
Here we're a little stuck. We want to be able to square $|a_n|$ and $|b_n|$ and bring them into their respective square-rooted sums. So if we label $a = \sqrt{\sum_{j=1}^{n-1} |a_j|^2}$, $b = \sqrt{\sum_{j=1}^{n-1} |b_j|^2}$, $c = |a_n|$, and $d = |b_n|$, we want to be able to say $ab + cd \leq \sqrt{a^2 + c^2} \sqrt{b^2 + d^2}$. In fact, we can say it! This inequality is always true for any $a, b, c, d \in \mathbb{R}$, because
$0 \leq (ad - bc)^2 = a^2 d^2 - 2abcd + b^2 c^2$
$\Rightarrow 2abcd \leq a^2 d^2 + b^2 c^2$
$\Rightarrow a^2 b^2 + 2abcd + c^2 d^2 \leq a^2 b^2 + a^2 d^2 + b^2 c^2 + c^2 d^2$
$\Rightarrow (ab + cd)^2 \leq (a^2 + c^2)(b^2 + d^2),$
and since both sides are positive reals, we can take the square root.
We now use this inequality to obtain
$|\sum_{j=1}^n a_j \overline{b_j}| \leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2} \sqrt{\sum_{j=1}^{n-1} |b_j|^2} + |a_n| |b_n|$
$\leq \sqrt{\sum_{j=1}^{n-1} |a_j|^2 + |a_n|^2} \sqrt{\sum_{j=1}^{n-1} |b_j|^2 + |b_n|^2}$
$= \sqrt{\sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2},$
and just square both sides to complete the inductive step.