Show that $f(z)=\sum_{k=1}^{n}|f_{k}(z)|$ cannot achieve a strictly local maximum inside a connected open set $D$. Suppose that we have $f_{1},\cdots, f_{n}$ holomorphic in a connected open set $D$. For $$f(z):=|f_{1}(z)|+|f_{2}(z)|+\cdot+|f_{n}(z)|\ \ \text{for}\ \ z\in D.$$ I want to show that $f(z)$ cannot achieve a strict local maximum inside $D$.

I know that this must have something to do with the maximum modulus principle, especially we have some connected open set, this will enable us to directly conclude that a function $g(z)$ is constant all over $D$ if it is holomorphic in $D$ and $|g|$ achieves a local maximum at $z_{0}\in D$.
However, I cannot connect this theorem to $f(z)$ because even thought $f_{k}(z)$'s are holomorphic, the summation of moduli of them may not be holomorphic.
I know a way to show that

Let $D$ be a disc with radius $R$ and center $z_{0}$. Suppose $f_{k}(z)$'s are holomorphic on $D$ and continuous on $\partial D$ such that $f(z_{0})\geq z$ for all $z\in D\cup C$, then $f$ is constant in $D\cup C$.

I can prove this claim using Cauchy's Integral formula, but it is long. This lemma can certainly conclude the proof quickly, but I realized that I did not really use the "connectedness" of $D$.

Is there any way for me to drive a shorter proof using maximum modulus principle? Thank you!

Edit 1:
The post is duplicated, but I'd still like to post a relatively complete proof.
Thanks to Conrad, the following is a relatively complete proof:
Suppose $\omega$ is the local maximum of $f(z):=\sum_{k=1}^{n}|f_{k}(z)|$. Note that as $f_{k}(\omega)$ is a complex number, unless it is zero, we can represent (uniquely) as $$f_{k}(\omega)=|f_{k}(\omega)|e^{it_{k}},\ \ t_{k}\in [0,2\pi).$$ Take $\alpha_{k}:=e^{-itk}$, and if $f_{k}(\omega)=0$ for some $k$, we tae $\alpha_{k}:=1$.
Then, we have $$f_{k}(\omega)=\alpha_{k}^{-1}|f_{k}(\omega)|\ \text{with}\ |\alpha_{k}|=1.$$
Now, define $g(z):=\sum_{k=1}^{n}\alpha_{k}f_{k}(z).$ Then, we have \begin{align*}
|g(z)|\leq \sum_{k=1}^{n}|\alpha_{k}||f_{k}(z)|=\sum_{k=1}^{n}|f_{k}(z)|&=f(z)\\
&\leq \sum_{k=1}^{n}|f_{k}(\omega)|=f(\omega)\\
&=\sum_{k=1}^{n}f_{k}(\omega)\alpha_{k}=g(\omega)\leq|g(\omega)|\ \ \ (*).
\end{align*}
Hence, $\omega$ is also the point at which $|g|$ achieves local max. As $g$ is holomorphic and $D$ is connected open, by the global maximum modulus principle, we conclude that $g(z)=C$ for some constant $C$ and for all $z\in D.$
But then, $(*)$ implies that $$|C|\leq C\leq |C|\implies C=|C|\implies C\geq 0.$$
So we have $g(z)=g(\omega)\geq 0$. But then (*) implies that $$g(\omega)=g(z)=|g(z)|\leq f(z)\leq f(\omega)\leq g(\omega),$$ and thus we must have $$g(\omega)=g(z)=f(z)=f(\omega)=C\geq 0\ \ \text{for all}\ \ z\in D,$$ and thus the local maximum of $f$ cannot be strict.

I really appreciate the patient and inspiring help from Conrad.
 A: Thanks to Conrad's patient and inspiring help, I wrote the proof, as follows:
Suppose $\omega$ is the point at which local maximum of $f(z):=\sum_{k=1}^{n}|f_{k}(z)|$ is achieved. Note that as $f_{k}(\omega)$ is a complex number, unless it is zero, we can represent (uniquely) as $$f_{k}(\omega)=|f_{k}(\omega)|e^{it_{k}},\ \ t_{k}\in [0,2\pi).$$ Take $\alpha_{k}:=e^{-itk}$, and if $f_{k}(\omega)=0$ for some $k$, we tae $\alpha_{k}:=1$.
Then, we have $$f_{k}(\omega)=\alpha_{k}^{-1}|f_{k}(\omega)|\ \text{with}\ |\alpha_{k}|=1.$$
Now, define $g(z):=\sum_{k=1}^{n}\alpha_{k}f_{k}(z).$ Then, we have \begin{align*}
|g(z)|\leq \sum_{k=1}^{n}|\alpha_{k}||f_{k}(z)|=\sum_{k=1}^{n}|f_{k}(z)|&=f(z)\\
&\leq \sum_{k=1}^{n}|f_{k}(\omega)|=f(\omega)\\
&=\sum_{k=1}^{n}f_{k}(\omega)\alpha_{k}=g(\omega)\leq|g(\omega)|\ \ \ (*).
\end{align*}
Hence, $\omega$ is also the point at which $|g|$ achieves local max. As $g$ is holomorphic and $D$ is connected open, by the global maximum modulus principle, we conclude that $g(z)=C$ for some constant $C$ and for all $z\in D.$
But then, $(*)$ implies that $$|C|\leq C\leq |C|\implies C=|C|\implies C\geq 0.$$
So we have $g(z)=g(\omega)\geq 0$. But then (*) implies that $$g(\omega)=g(z)=|g(z)|\leq f(z)\leq f(\omega)\leq g(\omega),$$ and thus we must have $$g(\omega)=g(z)=f(z)=f(\omega)=C\geq 0\ \ \text{for all}\ \ z\in D,$$ and thus the local maximum of $f$ cannot be strict.
