= Solution
Apply the preceding result to $L=\Delta$, $u=w$ and $f=\Delta w$. Define the barrier
$$
v(x)=\frac{|x|^2-d^2}{2n}.
$$
Then $\Delta v=1$. Since $0\in\Omega$ and $d=\operatorname{diam}\Omega$, $|x|=|x-0|\leq d$ on $\overline\Omega$, so $v\leq0$ on the boundary. All hypotheses of the barrier comparison hold, giving
$$
\boxed{w(x)\geq\frac1{2n}\left(\sup_\Omega(\Delta w)^+\right)(|x|^2-d^2)+\inf_{\partial\Omega}w.}
$$
The assumption that the origin belongs to the domain is what validates this particular choice of centered quadratic barrier.
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