Prove that: $\sup_{z \in \overline{D}} |f(z)|=\sup_{z \in \Gamma} |f(z)|$ Suppose $D=\Delta^n(a,r)=\Delta(a_1,r_1)\times \ldots  \times \Delta(a_n,r_n) \subset \mathbb{C}^n$
and 
$\Gamma =\partial \circ \Delta^n(a,r)=\left \{ z=(z_1, \ldots , z_n)\in \mathbb{C}^n:|z_j-a_j|=r_j,~ j=\overline{1,n}  \right \}$.
Let $f \in \mathcal{H}(D) \cap \mathcal{C}(\overline{D})$. 
Prove that: $\sup_{z \in \overline{D}} |f(z)|=\sup_{z \in \Gamma} |f(z)|$
I need your help. Thanks.

I'm not sure but I still post my solutions for the problem, if I have some mistakes, then you fix it. Thanks.
Case 1: $z\in D$
Since $D$ is bounded, its closure $\overline{D}$ is closed and bounded, hence compact.
$|f(z)|$ is continuous real-valued function on the compact set and so $\sup|f(z)|<\infty$. By the maximum modulus principle, we have $f$ is constant.
Because $f \in \mathcal{C}(\overline{D})$, therefore:$\sup_{z \in \overline{D}} |f(z)|=\sup_{z \in \Gamma} |f(z)|$.
Case 2: $z \in \Gamma$, it's obvious.
Case3: $z \in \partial D \setminus \Gamma$, 
We assume that $z_{1}^{0}, \ldots,z_{n}^{0}$ such that: $|z_j-a_j|=r_j, \forall j=1,\ldots, n_0$ and $z_{n_0+1}^{0},\ldots,z_{n}^{0}:|z_j-a_j|<r_j, \forall j=n_0+1,\ldots, n$.
We consider: $g :D' \to \mathbb{C} $, with $D'=\Delta(a_{n_0+1},r_{n_0+1}) \times \ldots \times \Delta(a_{n},r_{n})$ and $g$ is defined by
$z'=(z_{n_0+1}, \ldots , z_n) \mapsto g(z')=f(z_{1}^{0}, \ldots,z_{n}^{0},z_{n_0+1}, \ldots, z_n)$, so $g \in \mathcal{H}(D') \cap \mathcal{C}(\overline{D'})$.
If $g$ has $\sup$ at $z'^0=(z_{n_0+1}^{0},\ldots,z_{n}^{0}) \in D'$. By the maximum modulus principle, we have $g$ is constant in $D'$.
We have: 
$\sup_{z\in \overline{D}}|f(z)|=|f(z_0)|=|f(z_{1}^{0}, \ldots,z_{n}^{0})|=
|g(z'^0)|=|g(z^0)|$.
Where $z^*=(z_{n_0+1}^{*},\ldots,z_{n}{*}) \in \Gamma'$
$\implies \sup_{z\in \overline{D}}|f(z)|=|f(z_{1}^{0}, \ldots,z_{n}^{0},z_{n_0+1}^{*},\ldots,z_{n}{*})|=\sup_{z \in \Gamma} |f(z)|$.
Is that right? Thanks.
