Let $\Omega\subset\mathbb{R}^3$ be a bounded Lipschitz domain. Define the Hilbert space $$ H(div;\Omega):=\{u\in (L^2(\Omega))^3:\nabla\cdot u\in L^2(\Omega)\} $$ equipped with the graph norm $$ \|u\|_{H(div;\Omega)}:=\big(\|u\|^2_{(L^2(\Omega))^3}+\|\nabla\cdot u\|^2_{L^2(\Omega)}\big)^{1/2} $$ I have to prove that $$ \big|\int_{\partial\Omega}(u\cdot\nu)\phi\mathrm{d}A\big|\le\|u\|_{H(div;\Omega)}\|\phi\|_{H^1(\Omega)} $$ where $u\in(C^{\infty}(\bar{\Omega}))^3$,$\phi\in H^1(\Omega)$ and $\nu$ is the outward unitary normal vector on $\partial\Omega$.
The book I'm using says that the equality above is the Cauchy-Schwarz inequality, but I don't think so. I'm trying to prove it, but I can't.
Any help?
By divergence theorem $$ \int_{\partial \Omega }u\cdot n \ \phi=\int_\Omega div(u\phi) = \int_\Omega \phi div(u) + \nabla \phi \cdot u. $$ This gives $$ \left|\int_{\partial \Omega }u\cdot n \ \phi\right|\le \|\phi\|_{L^2(\Omega)}\|div(u)\|_{L^2(\Omega)} + \|\nabla \phi\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \le \|\phi\|_{H^1(\Omega)}\|u\|_{H(div,\Omega)}. $$