Junior Macedonian Mathematical Olympiad

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 $$\frac{\sqrt{a}}{2}+\frac{1}{1+a}=\frac{1}{2\sqrt{bc}}+\frac{1}{1+a}\ge \frac{1}{1+bc}+\frac{1}{1+a}=\frac{1}{1+\frac{1}{a}}+\frac{1}{1+a}=1\dots (1).$$ In the same way we prove that $$\frac{\sqrt{b}}{2}+\frac{1}{1+b}\ge 1\dots (2)$$ and $$\frac{\sqrt{c}}{2}+\frac{1}{1+c}\ge 1\dots (3).$$

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 $c=1$.