8. Let non-negative real numbers $a, b, c$ satisfy $a b+b c+c a=a+b+c>0$, then the minimum value of $\sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ is
A. 2
B. 3
C. $\sqrt{3}$
D. $2 \sqrt{2}$
$$
\begin{array}{l}
(a+b+c)(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) \\
=(a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a}+c \sqrt{a b}+a \sqrt{b c}+b \sqrt{c a} \\
\geqslant(a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a} \\
\geqslant 2(a b+b c+c a)=2(a+b+c) \Rightarrow \sqrt{a b}+\sqrt{b c}+\sqrt{c a} \geqslant 2,
\end{array}
$$

Equality holds when 2 of $a, b, c$ are 2 and 1 is 0. Therefore, the answer is $A$.