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.
Best Answer
Applying 1.35 with $a_1,\dots,a_n$ and $b_1,\dots,b_n$ being real numbers (which are included in complex numbers), we get $$ \left|\sum_{j=0}^{n}a_j {b_j}\right|^2\le\sum_{j=0}^{n}|a_j|^2\sum_{j=0}^{n}|b_j|^2 .$$
Can you take it from here?