Problem
Let $D\subset \mathbb{C}$ be a domain and $C$ its boundary with positive orientation with respect to D. Let $f:\overline{D} \rightarrow \mathbb{C}$ be a function which is continuous on $D$ and analytic in $\overline{D}$ such that
$$\int_{C} f(z)dz =M$$
Let $z_0\in D$ be a point such that {$z\in \mathbb{C}:\left|z-z_0\right|\leq \delta$}$\subset D$. Then, prove that $\left|f(z_0)\right|\leq\frac{\left|M\right|}{2\pi \delta}$.
My proof
I have, by the Cauchy's integral formula,
$$\left|f(z_0)\right|\leq\frac{1}{2\pi}\left|\int_{C_0}\frac{f(z)}{z-z_0}dz\right|\leq\frac{1}{2\pi}\int_{C_0}\left|\frac{f(z)}{z-z_0}\right|dz\leq\frac{1}{2\pi\delta}\int_{C_0}\left|f(z)\right|dz $$
where $C_0=${$z\in \mathbb{C}:\left|z-z_0\right|= \delta$} with positive orientation
If I can prove
$$\int_{C_0}\left|f(z)\right|dz\leq\left|\int_{C}f(z)dz\right|$$
, the problem solving is over. But I don't even know if this is possible. I think it might be possible if C is included in D, but it is not.
Is there a way to continue? Or should I look for another way?