Today we had this little problem in our FuncAna-Mock-Exam. I did not know any way to start this... I am thankfull for every hint and answer :)
As usual, $U \subset \mathbb{R}^d$ is open and bounded. Furthermore let $u\in W^{1,2}(U)$ satisfy $\int \sum_{j=1}^{d} \partial_j u \partial_jv \mathrm{d}m^d \leq 0$ for all $v\in W_0^{1,2}(U)$ with $v\geq0$ almost everywhere and $u^+ \in W_0^{1,2}(U)$. We have to show that $u \leq 0$ almost everywhere.
We know that $u^+$ satisfies $u^+ \ge 0$ a.e. and $u^+ \in W^{1,2}_0$, so we can use it as a "test function" in the inequality: $$ \int_\Omega \nabla u \cdot \nabla v = \int_\Omega \sum_{i=1}^d \partial_i u \partial_i v \le 0. $$ Using $v = u^+$ and noting that $\nabla u^+ = \nabla u \chi_{\{u >0\}}$, we find that $$ \int_{\{u >0\}} |\nabla u^+|^2 = \int_\Omega \nabla u \cdot \nabla u^+ \le 0. $$ Hence $\nabla u^+ =0$ a.e. in $\Omega$, and this tells us that $u^+$ is almost everywhere a constant (in each connected component of $\Omega$). However, since $u^+ \in W^{1,2}_0$ we know that actually this constant must be $0$ in each component, and so $u^+ = 0$ a.e. in all of $\Omega$. In turn this tells us that $u \le 0$ a.e. in $\Omega$.