Let $f(z)$ be an analytic function in the unit disc,and let $\sum_{n=0}^{\infty}a_nz^n$ be its Taylor expansion.Let $0<r<1$ and let $a_n=\alpha_n+i\beta_n$.Prove that for $n\ge1$, $$\alpha_n=\frac{1}{\pi}\int_0^{2\pi}u(r,\theta)\cos(n\theta)d\theta$$ $$\beta_n=\frac{1}{\pi}\int_0^{2\pi}u(r,\theta)\sin(n\theta)d\theta$$ where $u=Re\ f$.
Then let $A(r)=max_{|z|=r}Re\ f(z)$.Prove that $$|a_n|r^n\le max\{4A(r),0\}-2Re\ f(0)$$for all $n\ge 1$.
The formulas for $\alpha_n,\beta_n$ need a factor of $r^n$ on the left. When corrected, they will follow from $$a_n r^n = \frac{1}{2\pi}\int_0^{2\pi} f(re^{i\theta}) e^{-in\theta}\,d\theta$$ by separation of the real and imaginary parts.
To estimate integrals of the form $\int_0^{2\pi}u(r,\theta)\cos(n\theta)d\theta$ in terms of $\max u$, it would help to have a nonnegative factor next to $u$. But we have $\cos(n\theta)$. there. To fix this, add $1$ to the cosine. $$ \int_0^{2\pi}u(r,\theta)\cos(n\theta)\,d\theta = \int_0^{2\pi}u(r,\theta)(1+\cos(n\theta))\,d\theta - \int_0^{2\pi}u(r,\theta)\,d\theta $$ The first integral is estimated using $(1+\cos(n\theta))\le 2$: $$\int_0^{2\pi}u(r,\theta)(1+\cos(n\theta))\,d\theta \le 4\pi \max(A(r),0)$$ (can't use negative $A(r)$, because $u(r,\theta)(1+\cos(n\theta)) \le 2A(r)$ would not be a valid inequality).
The second integral is evaluated using the mean value property: $$\int_0^{2\pi}u(r,\theta)\,d\theta = 2\pi \operatorname{Re}f(0)$$ In conclusion, $$\alpha_n\le 4\max(A(r),0)-2\operatorname{Re}f(0) \tag{1}$$ To upgrade (1) to an estimate for $|a_n|$, use the "rotate until it's positive" trick. Namely, apply (1) to $f(\omega z)$ where $\omega $ is a unimodular complex number chosen so that for the rotated function, $a_n$ is a positive number.