Let $\Omega$ be a bounded open subset of $\mathbb{R^3}$, and $f$ be in $L^2(\Omega)$ Does there exist a weak solution in $W^{1,2}_0(\Omega)$ 
Let $\Omega$ be a bounded open subset of $\mathbb{R^3}$, and $f$ be in $L^2(\Omega)$. Does there exist a weak solution in $W^{1,2}_0(\Omega)$ to the following equation: 
\begin{cases}
\Delta u+\dfrac{1}{1+u^{2}}=f & \mathrm{in}\;\Omega\\
u=0 & \mathrm{on\;}\partial\Omega
\end{cases}

Help me with some hints to start.
Thanks in advanced.

In fact the idea of guacho works:
Let $\varphi(v)\in W^{1,2}_0(\Omega)$ be a weak solution of
$$
\Delta u+\frac{1}{1+v^2}=f.
$$
Such a weak solution exists as this means for $\varphi(v)$ that
$$
-\int_\Omega \nabla\varphi(v)\cdot\nabla w\,dx= 
\int_{\Omega}\left(f-\frac{1}{1+v^2}\right)w\,dx \quad\text{for all $w\in W^{1,2}_0(\Omega)$}.
$$
And since $\ell(w)=-\int_{\Omega}\left(f-\frac{1}{1+v^2}\right)w\,dx$ is a bounded linear 
functional on $W^{1,2}_0(\Omega)$, then there exists a $\varphi(v)\in W^{1,2}_0(\Omega)$, 
such that 
$$
\ell(w)=\langle w,\varphi(v)\rangle_{W^{1,2}_0(\Omega)}=\int_\Omega \nabla\varphi(v)\cdot\nabla w\,dx.
$$ 
Don't forget that $W^{1,2}_0(\Omega)$ is a Hilbert space with inner product $\langle w,w'\rangle_{W^{1,2}_0(\Omega)}=\int_\Omega \nabla w\cdot\nabla w'\,dx$.
Also, $\|\varphi(v)\|_{W^{1,2}_0(\Omega)}=\|\ell\|\le \|f\|_{L^2}+\|1\|_{L^2}=M.$ Hence the
nonlinear functional $\varphi$ maps $L^2(\Omega)$ into 
$$
B=\{w\in {W^{1,2}_0(\Omega)}: \|w\|_{{W^{1,2}_0(\Omega)}}\le M\}.
$$
Now $B\subset \{u\in L^2(\Omega) : \|u\|_{L^2}\le N\}$, for some $N>0$, due to Poincaré inequality. In particular, $\varphi$ maps $B$ into $B$, and $\varphi[B]\subset B$. But $B$
is compact subset of $L^2(\Omega)$, due to Rellich compactness theorem. Hence, 
Schauder fixed point theorem guarantees a fixed point $u$ for $\varphi$. Clearly $u\in L^2(\Omega)$, but $u=\varphi(u)\in B\subset W^{1,2}_0(\Omega)$.
Ὅπερ ἔδει δεῖξαι (=quod erat demostrantum). 
