2. Find all positive real numbers $a, b, c \leq 1$ such that
$$
\min \left\{\sqrt{\frac{a b+1}{a b c}}, \sqrt{\frac{b c+1}{a b c}}, \sqrt{\frac{a c+1}{a b c}}\right\}=\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}
$$

Modified from Problem 4 of the 2013 USA Math Olympiad
$$
a=\frac{1}{1+r^{2}}, b=\frac{1}{1+s^{2}}, c=\frac{1}{1+t^{2}} .
$$

Also, WLOG, suppose $t=\min \{r, s, t\}$. The required equation can then be rewritten as
$$
\sqrt{\left(1+t^{2}\right)\left\{1+\left(1+r^{2}\right)\left(1+s^{2}\right)\right\}}=r+s+t .
$$

By Cauchy-Schwarz, $(r+s+t)^{2} \leq\left([r+s]^{2}+1\right)\left(1+t^{2}\right)$, and so we have
$$
\left(1+r^{2}\right)\left(1+s^{2}\right) \leq(r+s)^{2},
$$
which is equivalent to $(r s-1)^{2} \leq 0$. Only equality is true here; hence $r s=1$ and all preceding inequalities become equations, and so $t(r+s)=$ 1. Now, conversely, if $r s=1$ and $t(r+s)=1$, with $t=\frac{1}{r+s}$ still less than both $\frac{1}{r}=s$ and $\frac{1}{s}=r$, the condition of the problem is still satisfied. Therefore, the solutions are
$$
a=\frac{1}{1+r^{2}}, b=\frac{1}{1+\frac{1}{r^{2}}}, c=\frac{\left(r+\frac{1}{r}\right)^{2}}{1+\left(r+\frac{1}{r}\right)^{2}}
$$
and permutations of these.