Show: $\max_{|z|=R} \operatorname{Re}\left(z\frac{f'(z)}{f(z)}\right) \geq N $ Let $f$ be a holomorphic function defined in a neighbourhood of $\overline{D(0,R)}$ which has no zero on $\partial D(0,R).$ Let $N$ be number of zeros of $f$ in $D(0,R).$ 
Show: $\max_{|z|=R} \operatorname{Re}\left(z\dfrac{f^{\prime}(z)}{f(z)}\right) \geq N \ $
My attempt: $\begin{align}\int_{|z|=R}\operatorname{Re}\left(z\dfrac{f^{\prime}(z)}{f(z)} \right) dz+ i\int_{|z|=R}\operatorname{Im}\left(z\dfrac{f^{\prime}(z)}{f(z)} \right)dz\end{align}=2\pi i \sum^{n}_{i=1}\alpha_{i} a_i,$ where $a_i$ is zero of $f$ and $\alpha_i$ is the corresponding multiplicity and $\alpha_1+...+\alpha_n = N.$
Could anyone advise me on how to proceed from here? Thank you. 

Note that
$$N=\frac{1}{2\pi i}\int_{|z|=R}\frac{f'(z)}{f(z)}dz=\frac{1}{2\pi}\int_0^{2\pi}\frac{f'(Re^{i\theta})}{f(Re^{i\theta})}Re^{i\theta}d\theta$$
