Minimal Value

Since $(a+b+c)^2\ge 3(ab+bc+ca)\ge 9$ hence $a+b+c\ge3$ .

We will now use Holder Inequality

$\left( \sum_{cyc}\frac{a}{\sqrt{b+c}} \right)^2\left( \sum_{cyc}a(b+c) \right) \ge (a+b+c)^3= (a+b+c)(a+b+c)^2\ge 3\cdot 3(ab+bc+ca)$ on the left side $\left( \sum_{cyc}\frac{a}{\sqrt{b+c}} \right)^2\left( \sum_{cyc}a(b+c) \right)=\left( \sum_{cyc}\frac{a}{\sqrt{b+c}} \right)^2\cdot 2 \cdot (ab+bc+ca)$ so the term $ab+bc+ca$ cancels out

$\left( \sum_{cyc}\frac{a}{\sqrt{b+c}} \right)^2\ge \frac{9}{2}\Leftrightarrow \left( \sum_{cyc}\frac{a}{\sqrt{b+c}} \right)\ge \frac{3}{\sqrt{2}}$ so $mn=6$