= Solution
We apply the closed-bounded-domain version of the <Schauder fixed-point theorem> in the <Banach space> $L^2(\Omega)$. Write $m=|\Omega|$ and use $G,T,C_P$ from part (b)(ii). Choose
$$
a=C_P^2m^{1/2},\qquad b=C_P^2m^{1/4},\qquad
R=\max\{1,2a,4b^2\},
\qquad C=\{w\in L^2(\Omega):w\geq0\text{ a.e.},\ \|w\|_2\leq R\}.
$$
The set $C$ is nonempty, bounded, closed and a <convex set>. For closedness of its positivity condition, if $w_j\geq0$ and $w_j\to w$ in $L^2$, then pointwise $w^-\leq|w-w_j|$, whence $\|w^-\|_2\to0$.
The <weak maximum principle> already gives $Tw\geq0$. Our estimates give, for $w\in C$,
$$
\|Tw\|_2\leq C_P^2\bigl(m^{1/2}+m^{1/4}\|w\|_2^{1/2}\bigr)
\leq a+b\sqrt R\leq R/2+R/2=R.
$$
Thus $T(C)\subset C$.
To prove <continuity>, use the pointwise <square root> estimate $|\sqrt s-\sqrt t|^2\leq|s-t|$ for $s,t\geq0$. The <Cauchy-Schwarz inequality> then gives
$$
\|\sqrt w-\sqrt z\|_2^2\leq\|w-z\|_1\leq m^{1/2}\|w-z\|_2,
\qquad
\|Tw-Tz\|_2\leq C_P^2m^{1/4}\|w-z\|_2^{1/2}.
$$
Hence $T$ is a <continuous map> on $C$, even though the <square root> need not have a bounded <derivative> near zero.
Finally,
$$
\|\nabla Tw\|_2\leq C_P\bigl(m^{1/2}+m^{1/4}\sqrt R\bigr)
\qquad(w\in C).
$$
The <Rellich-Kondrachov theorem> says that $H_0^1(\Omega)$ embeds compactly into $L^2(\Omega)$ on a bounded smooth <domain>. For $n>2$, $2<2n/(n-2)$, and for $n=2$ every finite target exponent is available, so the required <compact embedding> holds in every dimension in the question. Thus $T(C)$ is <relatively compact> in $L^2(\Omega)$.
All hypotheses of the <Schauder fixed-point theorem> are met. It gives $u=Tu\in C$, and part (b)(ii) converts this <fixed point> into the required <weak solution>:
$$
\boxed{\text{There exists }u\in H_0^1(\Omega),\ u\geq0\text{ a.e., with }
\int_\Omega\nabla u\cdot\nabla v\,dx=\int_\Omega(1+\sqrt u)v\,dx
\quad\forall v\in H_0^1(\Omega).}
$$
Only existence is needed. The construction does not assume that $\Omega$ is small or that $T$ is a <contraction mapping>.
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