= Solution
On the real interval, $h(x)$ is real and $H(x)=h(x)^2\geq0$. Expanding and integrating the finite sum yields
$$
\sum_{r,s=1}^n\frac{a_ra_s}{r+s+1}=\int_0^1h(x)^2\,dx.
$$
Apply part (b)(iv) to the entire function $H=h^2$. Since $|H(e^{i\theta})|=|h(e^{i\theta})|^2$, <orthogonality of complex exponentials> gives
$$
\int_0^{2\pi}|h(e^{i\theta})|^2\,d\theta=\sum_{r,s}a_ra_s\int_0^{2\pi}e^{i(r-s)\theta}\,d\theta=2\pi\sum_{r=1}^n a_r^2.
$$
Consequently
$$
\boxed{\left|\sum_{r,s=1}^n\frac{a_ra_s}{r+s+1}\right|\leq\pi\sum_{r=1}^n a_r^2.}
$$
The quadratic form is actually nonnegative, so its absolute value can be dropped. This <shifted Hilbert quadratic-form bound> comes from combining the logarithmic contour estimate with a finite <Parseval identity>.
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