We know the standard trace inequality: for a bounded domain with certain boundary regularity, there is a $C>0$ such that
$$ \|Tu\|_{L^2(\partial \Omega)}\leq C\|u\|_{H^1(\Omega)}, \quad \quad u\in H^1(\Omega). $$
Here $Tu$ is the trace of $u$ on $\partial \Omega$. I roughly remembered that I encountered a more convenient version of this trace inequality with $\epsilon$ somewher. Namely, for any given $\epsilon>0$, it holds
$$ \|Tu\|_{L^2(\partial \Omega)}\leq \epsilon \|\nabla u\|_{L^2(\Omega)}+C(\epsilon)\|u\|_{L^2(\Omega)}, \quad \quad u\in H^1(\Omega). $$
Anyone can provide an elementary proof for this? Highly appreciated!