Prove inequality $|u| \leq \max_{\partial{\Omega}}{|f|} + \frac{1}{4} \max_{\Omega}{|F|} \ R^2$ Consider the problem 
   $$\cases{ -\Delta{u}=F, \ in \ \Omega  \\ u=f \  \ ,on \ \partial{\Omega} }   $$ 
   $$$$
The maximum principle says that if $u$ is a solution of $-\Delta{u}=F$ with $F\leq0$, where $\Delta{u}$ is the Laplacian of $u$, then $u$ attains its maximum on the boundary $\partial{\Omega}$, that is $u\leq \max{f} $. 
Let $u$ be a solution of the differential equation and consider the function $v = u + \frac{1}{4}\max_{\Omega}{|F|} \ (x^2+y^2) $,   Note that 
$$ -\Delta{v} = -\Delta[u + \frac{1}{4}\max_{\Omega}{|F|} \ (x^2+y^2) ] = -\Delta{u}- \frac{1}{4}\max_{\Omega}{|F|} \ \Delta{(x^2+y^2)}   = F - \max_{\Omega}{|F|} \leq 0 $$
then $$v = u + \frac{1}{4}\max_{\Omega}{|F|} \ (x^2+y^2) \leq \max_{\partial{\Omega}}{f} + \frac{1}{4} \max_{\Omega}{|F|} \ R^2 \leq \max_{\partial{\Omega}}{|f|} + \frac{1}{4} \max_{\Omega}{|F|} \ R^2 $$ 
where $R$ is the radius of a circle containing $\Omega$( $\Omega$ is bounded ). Since $u \leq v$, we have that 
$$ u \leq \max_{\partial{\Omega}}{|f|} + \frac{1}{4} \max_{\Omega}{|F|} \ R^2   $$
Now I want to prove that 
$$ -( \max_{\partial{\Omega}}{|f|} + \frac{1}{4} \max_{\Omega}{|F|} \ R^2 ) \leq u  $$
to conclude that 
$$ |u| \leq \max_{\partial{\Omega}}{|f|} + \frac{1}{4} \max_{\Omega}{|F|} \ R^2  $$
But I don't know how to do that. 
 A: You have the right idea, but you need to be a little more careful to deal with the positive and negative parts of $F$ and $f$ separately.
I'm going to make the assumptions that $\Omega\subset\mathbb{R}^n$ is bounded and open, and that $F\in C_b(\Omega),f\in C(\partial\Omega)$. For notational convenience let $h^+=\max(h,0)$ and $h^{-}=\max(-h,0)$ for a continuous function $h$. 
Now let $M^+=\sup_{\Omega}F^+$ and $M_{-}=\sup_{\Omega}F^-$. Now fix $x_0\in\Omega$ and let $v\in C^2(\Omega)\cap C(\overline{\Omega})$ be defined by 
$$v(x)=u(x) + \frac{|x-x_0|^2}{2n}M^+.$$ We compute 
$$-\Delta v = -\Delta u - \Delta\left(\frac{|x-x_0|^2}{2n}M^+\right)=-\Delta u - M^+ = F - M^+ \leq 0.$$
Thus, by the weak maximum principle 
$$u(x_0)=v(x_0)\leq\sup_{x\in\overline{\Omega}}v(x)=\sup_{x\in\partial\Omega}v(x)\leq\sup_{x\in\partial\Omega}u(x)+k(x_0)M^+\leq\sup_{\partial\Omega}f^++k(x_0)\sup_{\Omega}F^+,$$
where $k(x_0)=\sup_{x\in\partial\Omega}\frac{|x-x_0|^2}{2n}.$ 
We can repeat an identical argument with the function $w\in C^2(\Omega)\cap C(\overline{\Omega})$ defined by 
$$w(x)=u(x)-\frac{|x-x_0|^2}{2n}M_{-}$$ to find that 
$$u(x_0)\geq -k(x_0)\sup_{\Omega}F^--\sup_{\partial\Omega}f^-.$$
Combining these together, we see that 
$$-k(x_0)\sup_{\Omega}F^--\sup_{\partial\Omega}f^-\leq u(x_0)\leq\sup_{\partial\Omega}f^++k(x_0)\sup_{\Omega}F^+.$$
Now notice that for all $x_0\in\Omega$ that $k(x_0)\leq \frac{1}{2n}(\operatorname{diam}\Omega)^2$, and so by taking the supremum over $x_0$ in the above we conclude that 
$$\|u\|_{L^\infty(\Omega)}\leq\|f\|_{L^\infty(\partial\Omega)} + \frac{1}{2n}(\operatorname{diam}\Omega)^2\|F\|_{L^\infty(\Omega)},$$
which is exactly your desired result (when $n=2$).
