Let $Ω ⊂ R^d$ be a bounded domain for which the divergence theorem holds. Assume $u ∈ C^2(\bar{Ω} ), u = 0$ on $∂Ω$. Show that for every $ε > 0,$
$2\int_Ω |∇u(x)|^2 dx ≤ ε\int_Ω(Δu(x))^2 dx +\frac{1}{ε}\int_Ω u^2(x) dx.$
I tried using Divergence theorem for some $w=\epsilon u^2$, but no success. At least a hint would be appreciated.
Let $\phi=\frac1{\sqrt \epsilon} u$ and $\psi =\sqrt{\epsilon}\, \nabla^2 u$. Using the divergence theorem along with the condition that $u=0$ for $x\in \partial \Omega$, we find
$$\begin{align} \int_\Omega \left|\nabla u\right|^2\,dx&=-\int_\Omega u \nabla^2 u\,dx\\\\ &=-\int_\Omega \phi \psi \,dx \tag 1 \end{align}$$
Now, applying the Cauchy-Schwarz inequality to the right-hand side of $(1)$ reveals
$$\left(\int_\Omega \left|\nabla u\right|^2\,dx\right)^2\le \left(\int_\Omega \phi^2\,dx\right)\left(\int_\Omega \psi^2\,dx\right) \tag 2$$
Then, applying the AM-GM inequality to the right-hand side of $(2)$ yields
$$\left(\int_\Omega \phi^2\,dx\right)\left(\int_\Omega \psi^2\,dx\right)\le \frac14 \left(\int_\Omega \phi^2\,dx+\int_\Omega \psi^2\right)^2 \tag 3$$
whence taking square roots and using $(2)$, we obtain the coveted inequality
$$\int_\Omega |\nabla u^2|\,dx\le \frac12 \int_\Omega \left(\frac1{\epsilon}u^2+\epsilon(\nabla^2 u)^2\right)$$