Let, $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be an entire function and let $r$ be a positive real number. Then, which is(/are) correct?
(a) $\sum_{n=0}^{\infty}|a_{n}|^{2}r^{2n}\le \sup_{|z|=r} |f(z)|^{2}$.
(b) $\sup_{|z|=r} |f(z)|^{2}\le \sum_{n=0}^{\infty}|a_{n}|^{2}r^{2n}$.
(c) $\sum_{n=0}^{\infty}|a_{n}|^{2}r^{2n}\le \dfrac{1}{2\pi}\int_{0}^{2\pi}\bigl|f\bigl(re^{i\theta}\bigr)\bigr|^{2}d\theta$.
(d) $\sup_{|z|=r} |f(z)|^{2}\le \dfrac{1}{2\pi}\int_{0}^{2\pi}\bigl|f\bigl(re^{i\theta}\bigr)\bigr|^{2}d\theta$.
This problem appeared as the No.82 problem in CSIR-UGC(NET) Mathematical Sciences Exam Paper, from New Delhi IMS in 2014, see this pdf
I tried using Cauchy's inequality but I can't proceed further.
Instead of Cauchy's formula, you need Parseval's formula. $$ \sum_{n=0}^\infty |a_n|^2r^{2n} = \frac1{2\pi} \int_{t=0}^{2\pi}\big|f(r^{it})\big|^2 \mathrm{d}t. $$
So we have equality in (c) and the trivial estimate $\frac1{2\pi} \int_{t=0}^{2\pi}\big|f(r^{it})\big|^2 \mathrm{d}t \le \sup_{|z|=r}|f(z)|^2$ in (a).
(b) and (d) are equivalent and false; try $f(z)=1+z$.