= Solution
The displayed nonlinearity is singular at $s=-1$, so it cannot be used directly as the globally bounded function in part (b). On $s\geq0$, however,
$$
\frac12\leq f(s)=\cosh\!\left(\frac1{1+s}\right)-\frac12\sin\sqrt{1+s^2}\leq\cosh(1)+\frac12.
$$
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 $F\geq1/2$. 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 $\int\nabla u\cdot\nabla\varphi\geq0$ for every nonnegative $\varphi\in H_0^1$, then $u\geq0$ <almost everywhere>. Here this also follows directly by testing with the negative part $u_-=\max\{-u,0\}\in H_0^1$:
$$
-\|\nabla u_-\|_2^2=\int_\Omega F(u)u_-\geq0.
$$
The <Poincaré inequality> gives $u_-=0$. Hence $u\geq0$, $F(u)=f(u)$, and the denominator $1+u$ stays at least one <almost everywhere>. Thus \b[the original equation has a nonnegative weak solution], with bounded right side.
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