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.
Let $v\in\mathbb{R}^p$ be a fixed vector. For $y\in\mathbb{R}^p$, let $\langle y,v\rangle$ denote the
standard inner product. The function $h_v:\mathbb{R}^p\to\mathbb{R}$ defined by
$$h_v(y)=\exp(\langle y,v\rangle )$$
is convex, because $e^t$ is convex. Indeed, if $0\leq\lambda\leq 1$, then for every $y_1,y_2\in\mathbb{R}^p$:
$$
h_v(\lambda y_1+(1-\lambda)y_2)=e^{\lambda\langle y_1,v\rangle+(1-\lambda)\langle y_2,v\rangle}\leq \lambda e^{\langle y_1,v\rangle}+(1-\lambda)e^{\langle y_2,v\rangle}=\lambda h_v(y_1)+(1-\lambda)h_v(y_2)
$$
by the convexity of $e^t$. For $1\leq k\leq p$, let $e_k$ denote the standard unit vectors in $\mathbb{R}^p$. Put
$$v_k=\sum_{i=1}^ke_i$$
Given positive numbers $b_i$, the functions $b_i h_{v_i}$ are convex for each $i$, and so also their sum. Note that
$$F(y)=\sum_{i=1}^pb_i\exp(\sum_{j\leq i}y_j)=\sum_{i=1}^p b_i h_{v_i}(y)$$
So $F$ is a convex function. Observe that the Hessian of $F$ coincides with with the Hessian of the given function $f$. Therefore, the function $f$ is also convex.
It can be shown that if $H_y$ is the Hessian of $f$ (or of $F$) evaluated at some point $y\in\mathbb{R}^p$ then
$$\langle H_y v,v\rangle=\exp(-\sum_{i=1}^py_i)\sum_{j=1}^pb_j\exp(\sum_{2\leq i\leq j}x_i)\left(\sum_{i=1}^jv_i\right)^2$$
which is seen to be positive for $v\neq 0$, but it is much easier to prove that $F$ is convex as above.
Best Answer
The inequality (2) is true. Moreover, it holds for any two nonempty subsets $A$ and $B$ of real numbers. The respective claim can be proved similarly to the proof of the first part of this answer. At the final step, suppose that we have $a$ numbers from $A$ and $b$ numbers from $B$ equal to $−1$ and $|A|-a$ numbers from $A$ and $|B|-b$ numbers from $B$ equal to $1$. Then the difference between the right-hand and the left-hand sides of the inequality is $$\frac{2(a+b)(|A|+|B|-a-b)}{|A|+|B|}-\frac{2a(|A|-a)}{|A|}-$$ $$\frac{2b(|B|-b)}{|B|}=\frac{2(|A|b-|B|a)^2}{|A|\cdot |B|\cdot |A+B|}\ge 0.$$