Let $\Omega \subseteq \mathbb R^n$ be a bounded, connected region. Furthermore, let $f: \Omega \to \mathbb R, g: \partial \Omega \to \mathbb R$ be continuous, and let $f$ be bounded. Finally, let $u \in C^2(\Omega, \mathbb R) \cap C(\overline \Omega, \mathbb R)$ be a solution of the equation
$$\begin{cases} \Delta u = f & \text{ on }\Omega \\ u \mid_{\partial \Omega} = g& \end{cases}$$
where $\Delta$ is the Laplacian. I now want to show that
$$\sup_{\overline \Omega} {|u|} \leq C \left( \sup_{\partial \Omega} |g| + \sup_{\Omega} |f| \right)$$
for some constant $C > 0$ that only depends on $\Omega$ (not on $f, g$ or $u$).
I'm given the hint that I might solve this by finding a $\lambda \in \mathbb R$ satisfying $\Delta (u + \frac{|x|^2}{2 n} \lambda) \geq 0$ on $\Omega$.
I must admit that I don't really know how to get started. I thought that once I actually found a $\lambda$ as in the hint, I probably need to use some of the theory for harmonic/subharmonic functions (maximum principle or Harnach's inequality maybe?), but I'm a bit lost on how to actually get to that point and how that would even help me to find such a constant $C$ for the function $u$ itself.
Let $M=\sup|f|$. We know that $\Delta u = f \ge -M$. How to make a subharmonic function out of this? Add another function $w$ whose Laplacian is $M$. Since $\Delta(|x|^2) = 2n$, the function $w(x) = M|x|^2/(2n)$ will do the job.
Letting $v = u+w$, we get $\Delta v\ge 0$, hence $\sup_{\Omega} v\le \sup_{\partial \Omega} v$. It remains to go back to $u$. On the left this is easy: $u\le v$. On the right use $v\le u+MD^2/(2n)$ where $D = \sup_{\partial\Omega}|x|$. Hence $$ \sup_{\Omega}u \le \sup_{\partial \Omega} u + \frac{MD^2}{2n} = \sup_{\partial \Omega} g + \frac{D^2}{2n} \sup_\Omega|f| $$ You can now choose $C$ so that the right hand side is bounded by $C(\sup_{\partial \Omega} |g| + \sup_\Omega |f|)$.
All this was an estimate of $\sup_{\Omega}u$, but the same applies to $\sup_{\Omega}(-u)$ by replacing $f$ and $g$ by $-f, -g$.