, 2015

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,
$$\begin{aligned} \sqrt{a} + \sqrt{b} + \sqrt{c} &\ge 2 - \frac{2}{1+a} + 2 - \frac{2}{1+b} + 2 - \frac{2}{1+c} \\ &= 6 - 2\left(\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}\right) = 2. \end{aligned}$$

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 $1$.