= Solution
Use the classical regularity setting $u\in C^2(D)\cap C(\overline D)$, with $D$ bounded. Continuity on the <compact set> $\overline D$ guarantees existence of a minimum. The inequality $\Delta u=f\leq0$ means that $u$ is <superharmonic function>.
A strict inequality $\Delta u<0$ would immediately exclude an interior minimum, because its <Hessian matrix> there would be positive semidefinite and have nonnegative <trace>. To handle equality, introduce the <quadratic perturbation proof of the minimum principle>:
$$
u_\varepsilon(x,y)=u(x,y)-\varepsilon(x^2+y^2),\qquad
\Delta u_\varepsilon=f-4\varepsilon<0.
$$
Its minimum must lie on $\partial D$. Choose $R$ so that $x^2+y^2\leq R^2$ on $\overline D$. For any $p\in D$,
$$
u_\varepsilon(p)\geq\min_{\partial D}u_\varepsilon
\geq\min_{\partial D}u-\varepsilon R^2.
$$
Letting $\varepsilon\downarrow0$ gives $u(p)\geq\min_{\partial D}u$. Therefore
$$
\boxed{\min_{\overline D}u=\min_{\partial D}u.}
$$
This asserts that a minimum is attained on the boundary, not that it cannot also be attained inside; a constant <function> attains it everywhere.
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