= Solution
Assume real bounded coefficients, with a principal part defining a <uniformly elliptic operator>, say $a^{ij}\xi_i\xi_j\geq\lambda|\xi|^2$ for some $\lambda>0$, and assume $c\leq0$. No smoothness of the boundary is needed for this classical <weak maximum principle for elliptic operators>. Its conclusion is
$$
\boxed{\sup_\Omega u\leq\max\{0,\sup_{\partial\Omega}u\}.}
$$
If $c=0$, the zero may be omitted. Only the symmetric part of the principal coefficients contributes to the <Hessian matrix> contraction.
For a strict inequality $Lu>0$, a positive interior maximum is impossible: there $Du=0$, $D^2u$ is negative semidefinite, and $cu\leq0$, so $Lu\leq0$. To reduce the non-strict case to this one, let $q(x)=e^{\kappa x_1}$. Boundedness of $b^1,c$ permits choosing $\kappa$ so large that
$$
Lq=q\bigl(\kappa^2a^{11}+\kappa b^1+c\bigr)
\geq q\bigl(\lambda\kappa^2-\|b^1\|_\infty\kappa-\|c\|_\infty\bigr)>0.
$$
Thus $L(u+\varepsilon q)>0$. The preceding maximum argument and continuity on the compact closure give
$$
\sup_\Omega(u+\varepsilon q)\leq\max\{0,\sup_{\partial\Omega}(u+\varepsilon q)\}.
$$
Let $\varepsilon\downarrow0$; the exponential is bounded on the bounded domain. This is the <exponential perturbation proof of the weak maximum principle with drift>, with the zeroth-order term included. For $c=0$, any interior maximum, of either sign, is impossible for the perturbed function, giving the sharper boundary bound. The sign restriction on $c$ is essential: a positive first <Dirichlet Laplacian eigenfunction> violates the conclusion for $L=\Delta+\lambda_1$.
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