= Solution
Because $|h^2|=|h|^2$, the <Parseval identity> for the <Taylor series> gives
$$
M(r):=\frac1{2\pi}\int_0^{2\pi}|h^2(re^{i\theta})|d\theta=\boxed{\sum_{n\geq1}|c_n|^2r^{2n}}.
$$
Differentiate this <power series> for $r<1$ and use <1/b/iii>:
$$
M'(r)=\frac2r\sum_{n\geq1}n|c_n|^2r^{2n}=\frac{2\operatorname{Area}(h(r\mathbb D))}{\pi r}\leq\frac{2r}{(1-r^2)^2}.
$$
Since $M(0)=0$, integration yields the <integral mean bound for an odd square-root transform>,
$$
\boxed{M(r)\leq\int_0^r\frac{2s}{(1-s^2)^2}ds=\frac{r^2}{1-r^2}}.
$$
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