 Solution

The displayed nonlinearity is singular at $s=-1$, so it cannot be used directly as the globally bounded function in part (b). On $s\ge 0$, however,Use nonnegative truncation of a semilinear source: define $F(s)=f(\max \{s,0\})$. This is continuous and bounded on all of $\mathbb{R}$, with $F\ge 1/2$. Part (b) gives a weak solution of $-\Delta u=F(u)$, $u\in H_0^1$.

The weak maximum principle in the needed form says that if $u\in H_0^1$ and $\int \nabla u\cdot \nabla \phi \ge 0$ for every nonnegative $\phi \in H_0^1$, then $u\ge 0$ almost everywhere. Here this also follows directly by testing with the negative part $u_{-}=\max \{-u,0\}\in H_0^1$:The Poincaré inequality gives $u_{-}=0$. Hence $u\ge 0$, $F(u)=f(u)$, and the denominator $1+u$ stays at least one almost everywhere. Thus the original equation has a nonnegative weak solution, with bounded right side.