Theorem:
If $a_1 , ... , a_n$ and $b_1 , ... , b_n$ are complex numbers, then
$$|\sum(a_jb_j^\ast)|^2\le\sum(|a_j|^2)\sum(|b_j|^2)$$
where $b_j^*$ is the conjugate of $b_j$ .
Put $A = \sum(|a_j|^2)$ , $B = \sum(|b_j|^2)$ , $C = \sum(a_jb_j^*)$
Proof:
To prove that inequality my book shows this equation:
$$\sum(|Ba_j - Cb_j|^2) = B(AB - |C|^2)$$
My question is how can I obtain the LHS of the equation?
I don't know if I understood your question correctly. Actually, I don't know if you are asking why does $\sum_{j=1}^{n}(|Ba_{j}-Cb_{j}|^{2})=B(AB-|C|^{2})$ hold or how can one obtain Schwartz inequality from it. I'm going to adress both problems.
Let $\langle x,y \rangle := x^{*}y$ denote the usual inner product in $\mathbb{C}$, so that $|z|^{2} \equiv z^{*}z = \langle z,z\rangle$. We have: \begin{eqnarray} |Ba_{j}-Cb_{j}|^{2} = \langle Ba_{j}^{*}-C^{*}b_{j}^{*},Ba_{j}-Cb_{j}\rangle = B^{2}|a_{j}|^{2}-BCa_{j}^{*}b_{j}-BC^{*}a_{j}b_{j}^{*}+|C|^{2}|b_{j}|^{2} \tag{1}\label{1} \end{eqnarray} Using (\ref{1}), we have also: \begin{align} \sum_{j=1}^{n}|Ba_{j}-Cb_{j}|^{2} &= B^{2}\sum_{j=1}^{n}|a_{j}|^{2}-BC\sum_{j=1}^{n}a_{j}^{*}b_{j}-BC^{*}\sum_{j=1}^{n}a_{j}b_{j}^{*}+|C|^{2}\sum_{j=1}^{n}|b_{j}|^{2} \\ &= B^{2}A-B|C|^{2}-B|C|^{2}+|C|^{2}B = B(BA-|C|^{2}) \tag{2}\label{2} \end{align} This proves your equation. Now, the other question is how to obtain Schwartz inequality from (\ref{2}). Well, note that $|Ba_{j}-Cb_{j}|^{2}\ge 0$ for every $j=1,...,n$, so that $\sum_{j=1}^{n}|Ba_{j}-Cb_{j}|^{2} \ge 0$. Thus, we must have: $$B(BA-|C|^{2})\ge 0 $$ But, by definition, $B\ge 0$, so that, the above inequality implies $BA-|C|^{2}\ge 0$. But this means $|C|^{2}\le BA$, which is the same as: \begin{eqnarray} |\sum_{j=1}^{n}a_{j}b_{j}^{*}|^{2} \le (\sum_{j=1}^{n}|a_{j}|^{2})(\sum_{j=1}^{n}|b_{j}|^{2}) \end{eqnarray} and this is Schwartz inequality.