Proving the Schwarz Inequality for Complex Numbers using Induction I want to prove the following version of the Schwarz Inequality for complex numbers $a_1, a_2, \ldots, a_n \in \mathbb{C}$ and $b_1, b_2, \ldots, b_n \in \mathbb{C}$:
$$|\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2$$
Rudin has a nice proof where he shows that
$$ \sum (|(\sum |b_j|^2)a_j - (\sum a_j \overline{b_j})b_j|^2) = (\sum |b_j|^2)(\sum |a_j|^2 \sum |b_j|^2 - |\sum a_j \overline{b_j}|^2) $$
Since the left side is $\geq 0$, so is the right side.  If $b_1 = b_2 = \ldots = b_n = 0$, the Schwarz inequality is trivial; otherwise, we have $\sum |b_j|^2 > 0$, so that $(\sum |a_j|^2 \sum |b_j|^2 - |\sum a_j \overline{b_j}|^2) \geq 0$.
However, I don't find this proof very intuitive (how the heck did he come up with it?!), and I'd prefer to try it by induction:  assume Schwarz holds for $n-1$, then show it's true for $n$.  Is it possible to do it this way?
Here's what I have so far:
Let $C = |\sum_{j=1}^{n-1} a_j \overline{b_j}|$.  Then
$$ |\sum_{j=1}^n a_j \overline{b_j}|^2 = |C + a_n \overline{b_n}|^2 = |C|^2 + 2|C||a_n \overline{b_n}| + |a_n \overline{b_n}|^2$$
Using the inductive hypothesis and some basic properties of the complex norm, we have
$$ |\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^{n-1} |a_j|^2 \sum_{j=1}^{n-1} |b_j|^2 + 2|C||a_n||b_n| + |a_n|^2 |b_n|^2$$
We could use the triangle inequality to separate out that $|C|$ in the middle, but I don't think that helps...
 A: 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.
