Since $(1-\sqrt{bc})^2 \ge 0$ it follows that $1+bc \ge 2\sqrt{bc}$, i.e. $\frac{1}{2\sqrt{bc}} \ge \frac{1}{1+bc}$. We get that
$$\frac{\sqrt{a}}{2} + \frac{1}{1+a} = \frac{1}{2\sqrt{bc}} + \frac{1}{1+a} \ge \frac{1}{1+bc} + \frac{1}{1+a} = \frac{1}{1+\frac{1}{a}} + \frac{1}{1+a} = 1\ldots(1).$$
In the same way we prove that
$$\frac{\sqrt{b}}{2} + \frac{1}{1+b} \ge 1\ldots(2)$$
and
$$\frac{\sqrt{c}}{2} + \frac{1}{1+c} \ge 1\ldots(3).$$

By adding (1), (2) and (3) we get the required inequality. Let us note that equality holds if and only if $1=\sqrt{bc}$, i.e. $a=1$. In the same way we get $b=1$ and $c=1$.