2015 Ninth STARS OF MATHEMATICS Competition

The condition in the statement is equivalent to $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$. If $t$ is a non-negative real number, then $\sqrt{t}\ge \frac{2t}{1+t}=2-\frac{2}{1+t}$, and equality holds if and only if $t$ is either $0$ or $1$. Consequently, $$\begin{array}{rl}\sqrt{a}+\sqrt{b}+\sqrt{c}& \ge 2-\frac{2}{1+a}+2-\frac{2}{1+b}+2-\frac{2}{1+c}\\ & =6-2\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)=2.\end{array}$$

By the preceding, equality holds if and only if one of the numbers $a$, $b$, $c$ is $0$, and the other two are both equal to $1$.