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.
We can prove that $a_k^{1-1/k} < a_k + \frac{2}{k} \sqrt{a_k}$.
Indeed, if $a_k \ge 1$, it is obvious;
and if $0 < a_k < 1$, by Bernoulli inequality $(1+x)^r \le 1 + rx$ for $0 < r \le 1$ and $x > -1$, we have
$a_k^{1-1/k} = a_k (a_k^{-1/2})^{2/k} = a_k(1 + a_k^{-1/2} - 1)^{2/k}
\le a_k [1 + (a_k^{-1/2} - 1)\frac{2}{k}]
< a_k + \frac{2}{k}\sqrt{a_k}$.
Thus, by Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
\sum_{k=2}^n a_k^{1-1/k} &< \sum_{k=2}^n a_k + \sum_{k=2}^n
\frac{2}{k} \sqrt{a_k}\\
&= \sum_{k=2}^n a_k + \sqrt{\sum_{k=2}^n \frac{4}{k^2}}\sqrt{\sum_{k=2}^n a_k}\\
&= s + 2\sqrt{\sum_{k=1}^n \frac{1}{k^2} - 1}\ \sqrt{s}\\
&< s + 2\sqrt{s}
\end{align}
where we have used $\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$
to get $\sqrt{\sum_{k=1}^n \frac{1}{k^2} - 1} < \sqrt{\frac{\pi^2}{6} - 1} < 1$.
(Q. E. D.)
Best Answer
After some more thinking, I managed to figure it out:
SOLUTION:
I substitute $a_k^2=b_k$ and raise the entire inequality to the power of two giving me:
$$\left(\sum _{k=1}^nb_k\right)^2\le n\sum _{k=1}^nb_k^2$$
And since the left side of the inequality consists of the terms $b_k^2$ and the terms $2b_ib_j$, we can subtract n of the square terms from both sides of the inequality:
$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2+nb_2^2+nb_3^2+...+nb_n^2-b_1^2-b_2^2-b_3^2-...-b_n^2$$ $$\sum _{i<j}^{ }2b_ib_j\le nb_1^2-b_1^2+nb_2^2-b_2^2+nb_3^2-b_3^2+...+nb_n^2-b_n^2$$ $$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+\left(n-1\right)b_3^2+...+\left(n-1\right)b_n^2$$ $$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)\sum _{k=1}^nb_k^2$$
We rearrange the inequality:
$$0\le \left(n-1\right)\sum _{k=1}^nb_k^2-\sum _{i<j}^{ }2b_ib_j$$
Which when expanded becomes:
$$0\le \left(n-1\right)\left(b_1^2+b_2^2+...b_n^2\right)-\left(2b_1b_2+2b_1b_3+....+2b_{n-1}b_n\right)$$ $$0\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+...\left(n-1\right)b_n-\left(2b_1b_2+2b_1b_3+....+2b_{n-1}b_n\right)$$ $$0\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+...+\left(n-1\right)b_n-2b_1b_2-2b_1b_3-....-2b_{n-1}b_n$$
To make a simple conclusion from this, I might just decide to see what happens when $n=3$. Then the inequality would become:
$$0\le 2b_1^2+2b_2^2+2b_3^2-2b_1b_2-2b_1b_3-2b_2b_3$$
Which is the same as:
$$0\le \left(b_1^2-2b_1b_2+b_2^2\right)+\left(b_1^2-2b_1b_3+b_3^2\right)+\left(b_2^2-2b_2b_3+b_3^2\right)$$ $$0\le \left(b_1-b_2\right)^2+\left(b_1-b_3\right)^2+\left(b_2-b_3\right)^2$$
And from here I would make the conclusion that:
$$0\le\sum_{i<j}^{ }\left(b_i-b_j\right)^2$$
And since this inequality is equivalent with the original inequality, we would then know that the original inequality is true.