Sharp constant in the curl inequality and ground states for curl-curl problem with critical exponent

Let $\Omega\subset\mathbb{R}^3$ be a Lipschitz domain and let $S_{\text{curl}}(\Omega)$ be the largest constant such that
$$\int_{\mathbb{R}^3}|\nabla\times u|^2dx\geq S_{\text{curl}}(\Omega)\inf_{w\in W_0^\sigma(\text{curl};\mathbb{R}^3)\\\nabla\times w=0}\left(\int_{\mathbb{R}^3}|u+w|^6dx\right)^{\frac{1}{3}}$$
for any $u$ in $W_0^6(\text{curl};\Omega)\subset W_0^6(\text{curl};\mathbb{R}^3)$ where $W_0^6(\text{curl};\Omega)$ is the closure of $C_0^\infty(\Omega, \mathbb{R}^3)$ in ${u \in L^6 (\Omega, \mathbb{R}^3):\nabla\times u \in L^2(\Omega, \mathbb{R}^3)}$ with respect to the norm $(|u|^2_6+|\nabla\times u|^2_2)^{1/2}$. We show that $S_{\text{curl}}(\Omega)$ is strictly larger than the classical Sobolev constant $S$ in $\mathbb{R}^3$ . Moreover, $S_{\text{curl}}(\Omega)$ is independent of $\Omega$ and is attained by a ground state solution to the curl-curl problem
$$\nabla\times(\nabla\times u)=|u|^4u$$
if $\Omega = \mathbb{R}^3$. With the aid of those results, we also investigate ground states of the Brezis-Nirenberg-type problem for the curl-curl operator in a bounded domain $\Omega$
$$\nabla\times(\nabla\times u)+\lambda u=|u|^4u \text{ }\text{ }\text{ in }\Omega$$
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with the so-called metallic boundary condition $\nu\times u=0$ on $\partial\Omega$ where $\nu$ is the exterior normal to $\partial\Omega$.