Proof of an inequality involving gradient function I'm reading ahead in my course, and I've encountered the following problem;

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain such that the
  divergence theorem holds. Assume that $u \in C^2(\bar{\Omega})$, $u =
0$ on $\partial \Omega$. Show that, $\forall \epsilon > 0$, we have;
  $$2 \int |\nabla u(x)|^2 dx \le \epsilon \int_{\Omega}(\Delta u(x))^2
 dx + \frac{1}{\epsilon} \int_{\Omega} u^2(x) dx$$

This problem looks really similar to the one here; A Cauchy with $\varepsilon$- type inequality for $C^1$ functions, but I'm unsure whether I can use similar logic within this problem. I haven't done analysis in a very long time, so I'm a bit unsure as to where I should begin with this problem. Any insights would be much appreciated!!
 A: Using the divergence theorem, 
$$2 \int_\Omega |\nabla u(x)|^2 dx = 2\int_\Omega \langle \nabla u, \nabla u\rangle dx =- 2\int_\Omega \text{div} \nabla u \cdot u dx + 2\int_{\partial\Omega} u \langle \nabla u, n\rangle dx$$ 
The last term is $0$ since $u$ is zero on the boundary. So (Note $\Delta u = \text{div}\nabla u$)
$$2 \int_\Omega |\nabla u(x)|^2 dx  \leq 2\int_\Omega |u \Delta u| dx$$ 
Now use the simple inequaity
$$2ab = 2(\sqrt\epsilon a)\left(\frac{b}{\sqrt\epsilon}\right) \leq  (\sqrt\epsilon a)^2 + \left(\frac{b}{\sqrt\epsilon}\right)^2$$
A: Start by using the divergence theorem and integration of parts.  Then,$$\int |\nabla u(x)|^2dx=\int \nabla u(x)\cdot \nabla u(x)dx=-\int u(x)\nabla^2 u(x)dx$$Note that the boundary integral vanishes since $u=0$ on the boundary.  Taking the squared-absolute value and applying Cauchy-Schwarz reveals$$\left|\int u(x)\nabla^2 u(x)dx\right|^2\le \int |\nabla^2 u(x)|^2 dx \int |u(x)|^2 dx$$Now, multiply the first term on the right by $\epsilon$ and dividing the second term on the right by $\epsilon$ gives $$\left|\int u(x)\nabla^2 u(x)dx\right|^2\le \epsilon\int |\nabla^2 u(x)|^2 dx \frac{1}{\epsilon}\int |u(x)|^2 dx$$Taking the square-root of both sides and applying the inequality of geometric and arithmetic means shows $$\left|\int u(x)\nabla^2 u(x)dx\right|\le\frac12\left[\epsilon\int |\nabla^2 u(x)|^2 dx +\frac{1}{\epsilon}\int |u(x)|^2 dx\right]$$
