Prove for elliptic PDE, $c<0$ implies $\sup_{\Omega} |u| \leq \sup_{\partial\Omega}\lvert \phi \rvert+ \sup_{\Omega}\lvert \frac{f}{c} \rvert.$

I have a problem that might be related to Alexandroff Maximum Principle. But I don't know how to prove it:

Assume $\Omega$ is a bounded domain. Let $u \in C^2(\Omega) \cap C^0(\overline{\Omega})$ satisfy $$Lu=f(x) \hspace{3mm} \text{in} \hspace{1mm} \Omega$$ $$u=\phi(x) \hspace{3mm} \text{on} \hspace{1mm} \partial\Omega$$ where $Lu=a^{ij}u_{ij}+b^iu_i+cu$. If $c=c(x)<0$, prove that $$\sup_{\Omega}\lvert u \rvert \leq \sup_{\partial\Omega}\lvert \phi \rvert+ \sup_{\Omega}\lvert \frac{f}{c} \rvert.$$

Here, for $Lu$, I'm using the Einstein notation, and $u_i := \frac{\partial u}{\partial x_i}$.

Anyone has any idea/hint on how to prove this?

This is exercise 3.7 in Chapter 3 (The Classical Maximum Principle) of

Gilbarg, David; Trudinger, Neil S., Elliptic partial differential equations of second order, Grundlehren der mathematischen Wissenschaften. 224. Berlin-Heidelberg-New York: Springer-Verlag. X, 401 p. (1977). ZBL0361.35003.

So this should follow by elementary means. Indeed the key idea is to use the fact that $c<0$ to deduce the non-existence of suitable interior max/min by considering the gradient and Hessian of $u$ at such a point. See below for a detailed proof along these lines.

Since $u$ is continuous on $\overline\Omega,$ it attains its supremum at some $x_0 \in \overline\Omega.$ Suppose that we have $$ u(x_0) > \sup_{\partial\Omega} \lvert\phi(x)\rvert + \sup_{\Omega} \left\lvert \frac{f}{c} \right\rvert. $$ Then since $u=\phi$ on $\partial\Omega,$ have $x_0 \in \Omega.$ Hence $x_0$ is an interior maximum, so if we consider $$ L_0u := \sum_{i,j} a^{ij}D_{ij}u + \sum_i b^iD_iu, $$ we have $L_0u(x_0) \leq 0.$ On the other hand since $c(x_0) < 0,$ by choice of $x_0$ we have $$ L_0u(x) = f(x_0) - c(x_0)u(x_0) > f(x_0) + |c(x_0)|\sup_{\Omega} \left\lvert \frac{f}{c} \right\rvert \geq 0, $$ which is a contradiction. The same argument applies to $-u,$ which implies the result.