$$\frac{1}{2a + \sqrt{bc}} = \frac{ab + bc + ca}{2a + \sqrt{bc}}.$$
Due to LM-GM, the inequality
$$ab + ca = a(b + c) \geq a \times 2\sqrt{bc}$$
holds. Hence
$$\frac{1}{2a + \sqrt{bc}} = \frac{ab + bc + ca}{2a + \sqrt{bc}} \geq \frac{2a\sqrt{bc} + bc}{2a + \sqrt{bc}} = \sqrt{bc},$$
and so
$$\sqrt{bc} + \frac{1}{2a + \sqrt{bc}} \geq 2\sqrt{bc}.$$
Similarly, it is deduced that
$$\sqrt{ca} + \frac{1}{2b + \sqrt{ca}} \geq 2\sqrt{ca},$$
and
$$\sqrt{ab} + \frac{1}{2c + \sqrt{ab}} \geq 2\sqrt{ab}.$$
Multiplying the latest three inequalities leads to the desired conclusion. ■