No, the answer is not $\frac{3}{\sqrt 2}$

We will prove that $\sqrt{\dfrac{b}{c + a}} \ge \dfrac{2b}{a + b + c}$

$\iff \dfrac{b}{c + a} \ge \dfrac{4b^2}{(a + b + c)^2}$

$\iff b(a + b + c)^2 \ge 4b^2(c + a)$

$\iff (a + b + c)^2 \ge 4b(c + a)$

$\iff \dfrac{(a + b + c)^2}{4} \ge b(c + a)$

$\iff \dfrac{a + b + c}{2} \ge \sqrt{b(c + a)}$ , which is totally true.

Similarly, $\sqrt{\dfrac{c}{a + b}} \ge \dfrac{2c}{a + b + c}$ and $\sqrt{\dfrac{a}{b + c}} \ge \dfrac{2a}{a + b + c}$ .

Plus all the above inequalities, we have that $\sqrt{\dfrac{a}{b + c}} + \sqrt{\dfrac{b}{c + a}} + \sqrt{\dfrac{c}{a + b}} \ge \dfrac{2b}{a + b + c} + \dfrac{2c}{a + b + c} + \dfrac{2a}{a + b + c} = \dfrac{2(a + b + c)}{a + b + c} = 2$ .

The equality happens when $0 \in \{a; \; b; \; c\}$ and others are equal one another.