I am doing the following problem.
Suppose $f(z)=\sum_{n=0}^{\infty}a_nz^n$ is an analytic function on the unit disc $|z|<1$. Let $0<r<1$. Prove that $$|a_n|r^n\leq \max\{4A(r),0\}-2Ref(0),$$ where $A(r)=\max_{|z|=r}Ref(z)$.
Here is my approach (but fail). Let $u(z)=Ref(z)$, then I have computed that (hope my computation is correct) $$Re(a_n)=\frac{1}{\pi r^n}\int_0^{2\pi}u(r,\theta)\cos(n\theta)d\theta.$$ $$Im(a_n)=\frac{1}{\pi r^n}\int_0^{2\pi}u(r,\theta)\sin(n\theta)d\theta.$$ Then, I used these to get the estimate \begin{align*} |a_n|r^n\leq&r^n(|Re(a_n)|+|Im(a_n)|) \\\leq&\frac{2}{\pi}\int_0^{2\pi}|u| \\=&\frac{2}{\pi}\bigg(\int_{u\geq 0}u-\int_{u\leq 0}u\bigg) \\=&\frac{2}{\pi}\bigg(2\int_{u\geq 0}u-\int_0^{2\pi}u\bigg) \\\leq&\frac{2}{\pi}\bigg(4\pi A(r)-\frac{2\pi}{2\pi}\int_0^{2\pi}u\bigg) \\=&8A(r)-4u(0), \end{align*} where I have used the mean value formula $$u(0)=\frac{1}{2\pi}\int_0^{2\pi}u(r,\theta)d\theta$$ for harmonic function. But it is seems that which is still over estimated (a factor of 2). Is the question wrong or there is a better way to obtain the required estimate? Thanks a lot!
One of the oldest tricks in the complex variables book is the identity $$|a|=\max_{\theta\in \mathbb R} \operatorname{Re}(e^{i\theta}a) \tag{1}$$ which holds for every complex number $a$. Thus, one can obtain an estimate of absolute value by estimating the real part of something, which has the advantage of being linear.
If the above hint is not enough, read further.