Write $f(z) = az^2 + bz + c$. We first prove that
$$|f(z)| \le 1 \text{ for all } z, |z| \le 1 \Leftrightarrow |f(z)| \le 1 \text{ for all } z, |z| = 1.$$

Assume that $f(z) = a(z - \alpha)(z - \beta)$. For any $z$, $|z| < 1$, if $\alpha = \beta$, one of the two intersection points of the line through $\alpha$ and the origin with the unit circle is closer to $\alpha$ than $z$ is; if $\alpha \neq \beta$, the line through $z$ and perpendicular to the line through $\alpha$, $\beta$ intersects the unit circle at two points, one of which is closer to $\alpha$, $\beta$ than $z$ is, respectively. So the equivalence holds.

For any complex number $z$, $|z| = 1$, it is obvious that
$$|f(z)| = |cz^{-2} + bz^{-1} + a|.$$
So $|ab|_{\max} = |bc|_{\max}$. Write $a'z^2 + b'z + c' = e^{i\alpha}f(e^{i\beta}z)$.
One can choose real numbers $\alpha$, $\beta$ such that $a'$, $b'$ are positive real numbers, so one can assume that $a$, $b \ge 0$ without loss of generality.
$$1 \ge |f(e^{i\theta})| \ge |\operatorname{Im} f(e^{i\theta})| = |a \sin 2\theta + b \sin \theta + \operatorname{Im} c|.$$
Without loss of generality we can assume $\operatorname{Im} c \ge 0$ (otherwise take a map, $\theta \to -\theta$). For any $\theta \in (0, \frac{\pi}{2})$,
$$\begin{align*} & 1 \ge a \sin 2\theta + b \sin \theta \ge 2\sqrt{ab} \sin 2\theta \sin \theta \\ \Rightarrow & ab \le \frac{1}{4 \sin 2\theta \sin \theta}, \quad \theta \in (0, \frac{\pi}{2}) \\ \Rightarrow & ab \le \min_{\theta \in (0, \frac{\pi}{2})} \frac{1}{4 \sin 2\theta \sin \theta} = \frac{1}{4 \max_{\theta \in (0, \frac{\pi}{2})} (\sin 2\theta \sin \theta)} = \frac{3\sqrt{3}}{16} \\ \Rightarrow & |bc|_{\max} = |ab|_{\max} \le \frac{3\sqrt{3}}{16}. \end{align*}$$

An example of $|bc| = \frac{3\sqrt{3}}{16}$ is
$$f(z) = \frac{\sqrt{2}}{8}z^2 - \frac{\sqrt{6}}{4}z - \frac{3\sqrt{2}}{8}, \\ |f(e^{i\theta})|^2 = 1 - \frac{3}{8} \left(\cos \theta - \frac{\sqrt{3}}{3}\right)^2 \le 1.$$