Hint: The quadratic equation $ax^2+bx+c=0$ has real roots if and only if $b^2-4ac\ge 0$.
Added: The quadratic polynomial $\sum_1^n (a_ix+b_i)^2$ is a sum of squares. Thus this polynomial is always $\ge 0$.
Recall that a quadratic $ax^2+bx+c$, where $a\gt 0$, is always $\ge 0$ if and only if the discriminant $b^2-4ac$ is $\le 0$. Compute the discriminant of the messy quadratic. The inequality $b^2-4ac\le 0$ turns out to be precisely the C-S Inequality (well, we have to divide by $4$).
As to when we have equality, the quadratic has a real root $k$ if and only if $a_ik+b_i=0$ for all $i$. This is the case iff $b_i=-ka_i$, meaning that the $a_i$ and $b_i$ are proportional.
By the way, things are I think marginally prettier if we look at the polynomial $\sum_1^n (a_ix-b_i)^2$.
If $\|a\| = \|b\| = 0$, then
\begin{align*}
0 & \leqslant \|a + b\|^2 =
\|a\|^2 + \|b\|^2 + 2\langle a, b \rangle =
+2\langle a, b \rangle, \\
0 & \leqslant \|a - b\|^2 =
\|a\|^2 + \|b\|^2 - 2\langle a, b \rangle = -2\langle a, b \rangle,
\end{align*}
therefore $\lvert\langle a, b \rangle\rvert = 0 \leqslant \|a\|\|b\|$.
Suppose, on the other hand, that $\|b\| > 0$. Then a fairly standard argument applies. Define $\lambda = \langle a, b \rangle/\|b\|^2$. Then
$\langle a - \lambda b, b \rangle = \langle a, b \rangle - \lambda \|b\|^2 = 0$, therefore
\begin{align*}
0 & \leqslant \langle a - \lambda b, a - \lambda b \rangle \\
& = \langle a, a - \lambda b \rangle - \lambda\langle b, a - \lambda b \rangle \\
& = \langle a, a - \lambda b \rangle - \lambda\langle a - \lambda b, b \rangle \\
& = \langle a, a - \lambda b \rangle \\
& = \|a\|^2 - \lambda \langle a, b \rangle,
\end{align*}
therefore
$$
\langle a, b \rangle^2 = \lambda\|b\|^2\langle a, b \rangle \leqslant \|a\|^2\|b\|^2,
$$
therefore
$$
\lvert\langle a, b \rangle\rvert \leqslant \|a\|\|b\|.
$$
The argument is similar when $\|a\| > 0$. So the Cauchy-Schwarz inequality holds in all cases.
Addendum
It appears (see my series of shame-faced comments below for details)
that this argument is merely an obfuscation of what is surely the
most "standard" of all proofs of the Cauchy-Schwarz inequality. It is
the one that is essentially due to Schwarz himself, and he had good
reasons for using it, quite probably including the fact that it makes
no use of the postulate that $\langle x, x \rangle = 0 \implies x = 0$!
In a modern abstract formulation, it goes as follows (assuming, of course,
that I haven't messed it up again). For all real $\lambda$, we have
$\|u\|^2 - 2\lambda\langle u, v \rangle + \lambda^2\|v\|^2 =
\|u - \lambda v\|^2 \geqslant 0$.
Therefore, the discriminant of this quadratic function of $\lambda$
must be $\leqslant 0$. That is, $\langle u, v \rangle^2 \leqslant \|u\|^2\|v\|^2$; equivalently, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.
Best Answer
No claim that $$\forall t\in\mathbb{R},\qquad f(t)>0$$ is ever made (as you note, this would indeed not be true for $t=0$). What is claimed is that $$\forall t\in\mathbb{R},\qquad f(t)\geq 0$$ which is true.
At a higher-level, why is the assumption that $\mathbf{b}\neq \textbf{0}$ is even made? It is only because for the rest of the proof we want $f(t)$ to be a degree-two polynomial (in $t$). Since $f(t) = \lVert \mathbf{b}\rVert^2 t^2 + 2t(\mathbf{a}\cdot\mathbf{b}) + \lVert \mathbf{a}\rVert^2$, this is only true of the coefficient if $t^2$ is non-zero.