I'm trying to solve a larger problem and have reduced it to showing that $$\int_\Omega (u-v) \Delta v dx \geq 0$$ Here $u, v$ are both continuous on $\overline\Omega$, $C^2$ on $\Omega$, which is an open region in $\mathbb{R}^n$ with smooth boundary. How can I prove this to be true?
EDIT: Additional information about the problem is that $u$ and $v$ agree on $\delta \Omega$ and $u$ is harmonic. I used this info to reduce my original problem (showing that $\int_\Omega |\nabla u|^2 \leq \int_\Omega |\nabla v|^2$) to the one I have now, but it might still be relevant.
Since $u$ is harmonic we can write
$$\int_{\Omega} (u-v) \nabla^2 v \,\,dV = -\int_{\Omega} (u-v) \nabla^2 (u-v) \,\,dV $$
Now, we use the vector identity $\phi \nabla^2 \psi=\nabla \cdot (\phi \nabla \psi)-\nabla \phi \cdot \nabla \psi$ with $\phi =\psi = u-v$ to reveal that
$$\begin{align} \int_{\Omega} (u-v) \nabla^2 v \,\,dV &= \int_{\Omega}\{ \nabla (u-v)\cdot \nabla (u-v)-\nabla \cdot \left((u-v) \nabla (v-u)\right) \}\,\,dV \\\\ &=\int_{\Omega} \left|\nabla (u-v)\right|^2\,\,dV-\int_{\partial \Omega}(u-v)\frac{\partial (u-v)}{\partial n} \,\,dS \end{align}$$
where we applied the divergence theorem to the second term on the right-hand side.
Note that $|\nabla (u-v)|^2\ge 0$, while the surface integral is zero since $u=v$ on $\partial \Omega$.
Thus, we arrive at
$$\begin{align} \int_{\Omega} (u-v) \nabla^2 v \,\,dV &=\int_{\Omega} \left|\nabla (u-v)\right|^2\,\,dV\\\\ &\ge 0 \end{align}$$
which was to be shown!
Note that the same inequality obtains if $\frac{\partial u}{\partial n}=\frac{\partial v}{\partial n}$,instead of $u=v$, on $\partial \Omega$.