= Solution
Use $h(\sqrt r\,e^{i\theta})^2=\phi(re^{2i\theta})$. As $\theta$ runs around the circle, the doubled angle goes around it twice, so <1/b/iv> gives
$$
\boxed{\frac1{2\pi}\int_0^{2\pi}|\phi(re^{i\theta})|d\theta=M(\sqrt r)\leq\frac r{1-r}}.
$$
The <Cauchy integral formula> for the <Taylor coefficients> now implies
$$
|a_n|\leq\frac1{2\pi r^n}\int_0^{2\pi}|\phi(re^{i\theta})|d\theta\leq\frac{r^{1-n}}{1-r}.
$$
For $n\geq2$, choosing $r=(n-1)/n$ gives the <linear coefficient bound for normalized univalent functions>,
$$
\boxed{|a_n|\leq n\left(1+\frac1{n-1}\right)^{n-1}\leq ne}.
$$
The last inequality follows from $\log(1+x)\leq x$ for $x>-1$.
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