Suppose $f$ analytic in the closed unit disk. Assume $\Re(f) \geq 0$ for all $|z| \leq 1$. Show the Taylor coefficients are $\Re(a_n) \leq 2$. 
Suppose $f$ analytic in the closed unit disk. Assume $\Re(f) \geq 0$ for all $|z| \leq 1$. If the Taylor expansion about the origin is
  $$
f(z) = 1 + a_1 z + a_2 z^2 + \cdots
$$
  then show the Taylor coefficients are $\Re(a_n) \leq 2$.

I've tried directly examining
$$
 2 - a_n 
$$
using the fact that
$$
a_n = \frac{1}{2 \pi i} \int_{|s| = 1} \frac{f(s)ds}{s^{n+1}}
$$
and looking at the modulus. 
I can also see that the Taylor expansion implies that $f(0)=1$ so there should exists points on the unit circle that contain real parts above and below $1$.
What other factors am I missing?
 A: Note first that \begin{align}
a_n=\frac{1}{2\pi i}\int_{|z|=1} \frac{f(z)}{z^{n+1}}dz=\frac{1}{2\pi}\int_0^{2\pi}f(e^{i\theta })e^{-in\theta }d\theta .\tag{1}
\end{align}
Especially
$$
2=2a_0=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta })d\theta $$
and hence we have \begin{align}
2=\frac{1}{\pi}\int_0^{2\pi} \Re (f(e^{i\theta }))d\theta .\tag{2}\end{align}
On the other hand, for $n\ge 1$ we have \begin{align}
0=\frac{1}{2\pi i } \int_{|z|=1}f(z)z^{n-1}dz=\frac{1}{2\pi}\int_0^{2\pi} f(e^{i\theta })e^{in\theta }d\theta .\tag{3}\end{align}
Therefore $(1)+(3)$ yields that $$
a_n=\frac{1}{2\pi}\int_0^{2\pi} f(e^{i\theta })\left(e^{-in\theta }+e^{in\theta }\right)d\theta
=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta })\cos n\theta d\theta
$$
and hence \begin{align}
\Re (a_n)=\frac{1}{\pi}\int_0^{2\pi} \Re (f(e^{i\theta }))\cdot \cos n\theta d\theta \tag{4}
\end{align}
Thus we see from (2) and (4) that $$
2-\Re (a_n)=\frac{1}{\pi}\int_0^{2\pi} \Re (f(e^{i\theta }))(1-\cos n\theta) d\theta\ge 0,
$$
since $\Re f\ge 0$.
