Prove: for $f,g \in Hol(G\subset \mathbb{C})$ $\implies$ max $(|f|+|g|)$ is on the boundary of $G$.
I don't really have a direction for this. I know it's got to do with the maximum modulus principle, but I'm not really sure how to get to it. Any guidance is welcome!
Assume that the maximum of $\lvert\,f\rvert+\lvert g\rvert$ is assumed in an interior point $z_0\in G$. Then, there are $a,b\in\mathbb C$, with $\lvert a\rvert=\lvert b\rvert=1 $, such that $$ \lvert\,f(z_0)\rvert+\lvert g(z_0)\rvert=af(z_0)+bg(z_0). $$ But then the harmonic function $$h(z)= \mathrm{Re} \big(af(z)+bg(z)\big),$$ assumes its maximum in an interior point, and hence it is constant, and thus so is $af(z)+bg(z)$. Let $$ af(z)+bg(z)=c. $$ Then $$ g(z)=\overline{b}\big(c-af(z)\big), $$ and $$ c=\lvert\,f(z_0)\rvert+\lvert g(z_0)\rvert\ge \lvert\,f(z)\rvert+\lvert g(z)\rvert=\lvert\,af(z)\rvert+\lvert c-af(z)\rvert.\tag{1} $$ But triangular inequality provides that $$c\le \lvert\,af(z)\rvert+\lvert c-af(z)\rvert.\tag{2} $$ But (1) and (2) imply that $$c= \lvert\,af(z)\rvert+\lvert c-af(z)\rvert. $$ This is true only if $af(z)$ is non-negative real, hence $f$ is constant and so is then $g$.
Therefore, if the maximum is achieved in the interior, then $f,g$ are constant.