Proof of Cauchy-Schwarz Inequality 
Here is the proof of the Cauchy-Schwarz inequality. I couldn't understand how the writer inferred that putting $x=-\frac{B}{A}$ in (1.24) proves the theorem, could you please expand it for me?
Thanks in advance.

What we want to prove is
$$
\left|\sum_{k=a}^n a_k b_k\right| \le \sqrt{\sum_{k=1}^n a_k^2 \sum_{k=1}^n b_k^2\quad{}},
$$
or equivalently
$$
\left( \sum_{k=a}^n a_k b_k \right)^2 \le \sum_{k=1}^n a_k^2 \sum_{k=1}^n b_k^2.
$$
We have
\begin{align}
A & = \sum_{k=1}^n a_k^2, \\[6pt]
B & = \sum_{k=1}^n a_k b_k, \\[6pt]
C & = \sum_{k=1}^n b_k^2.
\end{align}
So the goal is to prove $B^2\le AC$, with equality only if there is some scalar $x$ such that $x(a_1,\ldots,a_n)=(b_1,\ldots,b_n)$.
Notice that since $A=a_1^2+\cdots+a_n^2$, $A$ cannot be negative.  Then notice that
$$
\left( x + \frac B A \right)^2\qquad \begin{cases} =0 & \text{if }x=-B/A, \\[6pt]
> 0 & \text{if }x\ne -B/A. \end{cases}
$$
Consequently
$$
A x^2 + 2Bx + C \qquad\begin{cases} = \dfrac{B^2-AC} A & \text{if }x=-B/A, \\[6pt]
>\dfrac{B^2-AC} A & \text{if }x\ne B/A. \end{cases}
$$
Thus the smallest possible of $Ax^2+2Bx+C$ is $\dfrac{B^2-AC} A$.
The smallest possible value cannot be negative since $Ax^2+2Bx+C$ is a sum of squares:
$$
Ax^2+2Bx+C = (a_1 x+b_1)^2+\cdots+(a_n x+b_n)^2. \tag 1
$$
Hence $\dfrac{B^2-AC} A\le 0$.  Since $A>0$, this implies $B^2-AC\le 0$, which was to be proved.
We can have $B^2-AC=0$ only if for some $x$, the sum in $(1)$ is $0$.  But that can happen only if every term in the sum is $0$, and that implies that for some $x$ we have $x(a_1,\ldots,a_n)=(b_1,\ldots,b_n)$.
