We will prove first that $\sqrt{2+x} + \sqrt{y+z} = \sqrt{2+x} + \sqrt{4-x} \ge 2 + \sqrt{2}$.
Indeed, this is equivalent to
$$2 + x + 4 - x + \sqrt{(2 + x)(4 - x)} \ge 4 + 2 + 2\sqrt{2} \Leftrightarrow \\ x(2 - x) \ge 0,$$
which is true, since $x \in [0,2]$.
Similarly we have $\sqrt{2+y} + \sqrt{x+z} \ge 2 + \sqrt{2}$ and $\sqrt{2+z} + \sqrt{x+y} \ge 2 + \sqrt{2}$.
Adding all these, we get $A \ge 6 + 3\sqrt{2}$ and the equality holds, for example, if $x = y = 2$ and $z = 0$.