Solution:
Let $r, s, t \geq 0$ such that
$$a=\frac{1}{1+r^{2}},\ b=\frac{1}{1+s^{2}},\ c=\frac{1}{1+t^{2}}$$
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.