Chinese Mathematical Olympiad

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

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 $\alpha \ne \beta$, 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 $$|f(z)|=|c{z}^{-2}+b{z}^{-1}+a|.$$ So $|ab{|}_{\max }=|bc{|}_{\max }$. Write $a^{\prime }{z}^2+b^{\prime }z+c^{\prime }=e^{i\alpha }f(e^{i\beta }z)$. One can choose real numbers $\alpha$, $\beta$ such that $a^{\prime }$, $b^{\prime }$ are positive real numbers, so one can assume that $a$, $b\ge 0$ without loss of generality. $$1\ge |f(e^{i\theta })|\ge |\mathrm{Im}f(e^{i\theta })|=|a\sin 2\theta +b\sin \theta +\mathrm{Im}c|.$$ Without loss of generality we can assume $\mathrm{Im}c\ge 0$ (otherwise take a map, $\theta \to -\theta$). For any $\theta \in (0,\frac{\pi }{2})$, $$\begin{array}{rl}& 1\ge a\sin 2\theta +b\sin \theta \ge 2\sqrt{ab}\sin 2\theta \sin \theta \\ \Rightarrow & ab\le \frac{1}{4\sin 2\theta \sin \theta },\theta \in (0,\frac{\pi }{2})\\ \Rightarrow & ab\le \underset{\theta \in (0,\frac{\pi }{2})}{\min }\frac{1}{4\sin 2\theta \sin \theta }=\frac{1}{4{\max }_{\theta \in (0,\frac{\pi }{2})}(\sin 2\theta \sin \theta )}=\frac{3\sqrt{3}}{16}\\ \Rightarrow & |bc{|}_{\max }=|ab{|}_{\max }\le \frac{3\sqrt{3}}{16}.\end{array}$$

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