Set $x = \frac{b^2}{a^2}$, $y = \frac{c^2}{b^2}$, $z = \frac{a^2}{c^2}$, then $x, y, z \in \mathbb{R}^+$ and $xyz = 1$. It suffices to prove that
$$1 < \frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+y}} + \frac{1}{\sqrt{1+z}} \le \frac{3\sqrt{2}}{2}.$$

Without loss of generality, we assume that $x \le y \le z$. Set $A = xy$, we have $z = \frac{1}{A}$, $A \le 1$. Thus
$$\frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+y}} + \frac{1}{\sqrt{1+z}} > \frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+\frac{1}{x}}} = \frac{1+\sqrt{x}}{\sqrt{1+x}} > 1.$$
Let $u = \frac{1}{\sqrt{1+A+x+\frac{A}{x}}}$, then $u \in (0, \frac{1}{1+\sqrt{A}}]$, and $x = \sqrt{A}$ if and only if $u = \frac{1}{1+\sqrt{A}}$. Hence
$$\left( \frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+y}} \right)^2 = \left[ \frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+\frac{A}{x}}} \right]^2$$
$$= \frac{1}{1+x} + \frac{1}{1+\frac{A}{x}} + \frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\ = \frac{2+x+\frac{A}{x}}{1+A+x+\frac{A}{x}} + \frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\ = 1 + (1-A)u^2 + 2u.$$
Set $f(u) = (1-A)u^2 + 2u + 1$, we see that $f(u)$ is an increasing function on $u \in (0, \frac{1}{1+\sqrt{A}}]$, which implies that
$$\frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+y}} \le \sqrt{f\left(\frac{1}{1+\sqrt{A}}\right)} = \frac{2}{\sqrt{1+\sqrt{A}}}$$
Now set $\sqrt{A} = v$, we get
$$\frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+y}} + \frac{1}{\sqrt{1+z}} \le \frac{2}{\sqrt{1+\sqrt{A}}} + \frac{1}{\sqrt{1+\frac{1}{A}}} \\ = \frac{2}{\sqrt{1+v}} + \frac{\sqrt{2}v}{\sqrt{2}(1+v^2)} \le \frac{2}{\sqrt{1+v}} + \frac{\sqrt{2}v}{1+v} \\ = \frac{2}{\sqrt{1+v}} + \sqrt{2} - \frac{\sqrt{2}}{1+v} = -\sqrt{2} \left[ \frac{1}{\sqrt{1+v}} - \frac{\sqrt{2}}{2} \right]^2 + \frac{3\sqrt{2}}{2} \\ \le \frac{3\sqrt{2}}{2}.$$