# The number of zeros of a function regular on the closure on $U$ and the max of $\operatorname{Re}\frac{zf'(z)}{f(z)}$

Suppose that $f(z)$ is holomorphic on the closed unit disk $\bar U$ and never vanishes on the boundary $\partial \bar U$. Prove that the maximum of $\displaystyle \operatorname{Re}\frac{zf'(z)}{f(z)}\geq\#\mbox{zeros of$f(z)$in the unit disk.}$

Thoughts: By the argument principle$\displaystyle \int_{\partial \bar U}\frac{f'(z)}{f(z)}\mathrm{d}z=2\pi i\,\#\mbox{zeros}$. There is a similarity between the pattern of $\frac{f'}{f}$ and $\frac{zf'}{f}$, plus $\displaystyle\int \frac{zf'}{f}\mathrm{d}z=2\pi i\sum\mbox{roots}$ as a consequence of the argument principle. This might give us some ways. Another idea is to use the Borel-Caratheodory inequality.

I assume there is a typo in you question and the real question is how to show the maximum of $\Re\left[ \frac{zf'(z)}{f(z)}\right]$ over $\partial\bar{U}$ is greater than or equal to the number of zeros over $U$. Otherwise, your question is simply false.
In any event, you don't really need anything complicated. You just need to rewrite the formula for number of zeroes as an integral over $[0,2\pi]$. \begin{align} \#\text{zeroes} &= \frac{1}{2\pi i}\int_{\partial \bar{U}} \frac{f'(z)}{f(z)}dz = \frac{1}{2\pi}\int_0^{2\pi} \frac{e^{i\theta} f'(e^{i\theta})}{f(e^{i\theta})}d\theta = \frac{1}{2\pi}\int_0^{2\pi} \Re\left[\frac{e^{i\theta} f'(e^{i\theta})}{f(e^{i\theta})}\right]d\theta \\ &\le \max_{z\in\partial\bar{U}}\,\Re\left[ \frac{zf'(z)}{f(z)}\right] \end{align}