I've proved the following: Let $a_j$ and $b_j$ be sequences of non-negative numbers and for $k\geq 0$, $c_k=\sum_{j=0}^ka_jb_{k-j}$. Then $$\sum_{k=0}^\infty \cfrac{c_k^2}{k+1}\leq\left(\sum_{j=0}^\infty a_j^2\right)\left(\sum_{j=0}^\infty b_j^2\right)$$
Using this, I need to prove that for $f$ and $g$ (complex) analytic in the disk $\Delta(0,\rho)$, for $0<r<\rho$,
$$\cfrac{1}{\pi r^2}\int_{\Delta(0,r)}|f|^2|g|^2dA\leq\left(\cfrac{1}{2\pi}\int_0^{2\pi}|f(re^{it})|^2dt\right)\left(\cfrac{1}{2\pi}\int_0^{2\pi}|g(re^{it})|^2dt\right)$$
I'm not sure how to begin. In order to use the result above, I know I can expand the functions $f$ and $g$ as Taylor series in the disk $\Delta(0,r)$ but I don't understand how to evaluate/estimate the integral on the left hand side. Any help with this is appreciated. Thanks.
Let $f(z) = \sum_{n=0}^\infty a_n z^n$ and $g(z) = \sum_{n=0}^\infty b_n z^n$ be the power series of $f$ and $g$, respectively. Note that the $c_n$ are exactly the coefficients of the Cauchy product of these two power series, so we have $f(z) g(z) = \sum_{n=0}^\infty c_n z^n$.
Parseval's identity states that $$ \frac{1}{2\pi}\int_0^{2\pi}|f(re^{it})|^2 \, dt = \sum_{n=0}^\infty |a_n |^2r^{2n} $$ for $0<r<\rho$, and similarly for $g$ and $fg$.
Now compute the area integral with polar coordinates, apply Parseval's identity to the product $fg$, and integrate term-by-term: $$ \frac{1}{\pi r^2}\int_{\Delta(0,r)}|f|^2|g|^2 \,dA = \frac{1}{\pi r^2} \int_0^r \int_0^{2\pi}|(fg)(R e^{it})|^2 R \, dt \,dR \\ = \frac{1}{\pi r^2} \int_0^r 2 \pi\sum_{n=0}^\infty |c_n |^2R^{2n} R \, dR = \cdots = \sum_{n=0}^\infty \frac{|c_n |^2r^{2n}}{n+1} \, . $$
This shows that the desired inequality between the integrals is equivalent to the inequality $$ \sum_{n=0}^\infty \frac{|c_n |^2r^{2n}}{n+1} \le \left (\sum_{n=0}^\infty |a_n |^2r^{2n} \right) \left (\sum_{n=0}^\infty |b_n |^2r^{2n} \right) $$ and that is what you proved in the first part, applied to the sequences $(r^j a_j)_j$ and $(r^j b_j)_j$ .