Given that $f$ is a holomorphic function on the open unit disk where its real part is always positive ($\Re f(0) >0$), show that $|f^{'}(0)| \leq 2\Re (f(0))$.
My attempt: I used Cauchy's formula, the fact that the real part is positive and harmonic to obtain the following $$|f^{'}(0)| \leq 2 \Re f(0)+ 2\max_{|z|=1/2} \Im f(z)$$ However, I don't see a way to produce the desired sharper bound.
Context: Studying for Quals #8 in http://www.math.tamu.edu/graduate/phd/quals/ncomplex/j16.pdf
Define $\phi (z) = \frac{z-f(0)}{z+\overline{f(0)}}$ which takes values from ${\Re z > 0}$ to the open unit disk. Now we consider the function $\phi (f(z))$ which takes values from the open unit disk to itself. Also, $\phi (f(0))=0$. So, we may apply Schwarz's Lemma to obtain that $$|(\phi ( f(0)))^{'}| \leq 1.$$ Following through with the calculation, we obtain $$|(\phi ( f(0)))^{'}| = \frac{f^{'}(0)(f(0)+\overline{f(0)})}{(f(0)+\overline{f(0)})^2} \leq 1.$$ Since, $f(0)+\overline{f(0)} = 2 \Re f(0)$, we obtain $$|f^{'}(0)| \leq 2 \Re f(0)$$ as desired.