Suppose $f$ is holomorphic on the closed disc of radius $R$ around some point $z_0$ and $f(z) \neq 0$, i.e. $f$ is holomorphic on the disc $D_{R}(z_0)$, show that the minimum value of $f(z)$ on the disc occurs when $z$ is on the boundary of the disc.
I start with the Cauchy integration formula $$f(z_0) = \frac{1}{2\pi i}\int_{C_{R}(z_0)} \frac{f(z)}{z-z_0}\,dz$$
I then use the estimation lemma to show that $$|f(z_0)| =\frac{1}{2\pi i} |\int_{C_{R}(z_0)} \frac{f(z)}{z-z_0}\,dz| \leq \frac{1}{2\pi i} \max_{z \in C_{R}(z_0)} |\frac{f(z)}{z-z_0}|\cdot\ell(C_{R}(z_0))$$
Now we have $|z-z_0| = R$ for all $z \in C_{R}(z_0)$ and $\ell(D_{R}(z_0)) = 2\pi R$ meaning the above equation simplifies down to $$f(z_0)| \leq -i \cdot \max_{z \in C_{R}(z_0)} |f(z)|$$
This is where I get stuck, I am sure that I am closer to the solution but am not sure how to proceed.
EDIT: $f(z) \neq 0$ anywhere