Integral inequality (Divergence theorem) I'm trying to prove the following inequality:
$$2 \int_U |\nabla \phi|^2 dx \leq \int_U \phi^2 dx + \int_U |\Delta \phi|^2 dx$$
where $U \subset \mathbb{R}^n$ is bounded and open and $\phi \in C^\infty_c(U)$. I actually think I have managed to prove this just using one of Green's identities $$\int_UD \phi \cdot D \phi dx = -\int_U \phi \Delta \phi dx + \int_{\partial U}\frac{\partial \phi}{\partial \nu} \phi dx$$
(which comes from the divergence theorem), and then using the fact that $\phi = 0$ on $\partial U$. This then gives:
$$\int_U |\nabla \phi|^2 dx = -\int_U \phi \Delta \phi dx \leq \int_U |\phi| |\Delta \phi| dx \leq \int_U \frac{\phi^2}{2}+\frac{|\Delta \phi|^2}{2}$$ where the last inequality comes from Cauchy's inequality $ab \leq a^2/2 + b^2/2$. 
Does this seem correct? The problem is that I have been given a hint which says to use the fact that $\nabla \cdot(\phi \nabla \phi) = |\nabla \phi|^2 + \phi \Delta \phi$ and I'm not sure how to use this hint. Would this just lead to an alternative proof of the inequality?

You already used it :) That's where the Green's identity came from! 
