= Solution
\b[True.] This follows from the <maximum principle for harmonic functions>. A direct reduction to part (i) is also possible: for $\varepsilon>0$, set
$$
\phi_\varepsilon(x)=\phi(x)+\varepsilon|x|^2.
$$
Then $\Delta\phi_\varepsilon=4\varepsilon>0$, so
$$
\max_{\overline\Omega}\phi_\varepsilon
=\max_{\partial\Omega}\phi_\varepsilon.
$$
Because $\Omega$ is bounded, $\phi_\varepsilon\to\phi$ uniformly as $\varepsilon\downarrow0$. Taking the limit yields
$$
\boxed{\max_{\overline\Omega}\phi=\max_{\partial\Omega}\phi.}
$$
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