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) $2 \sqrt{2}$
8. A.

Assume $a \geqslant b \geqslant c$. By the AM-GM inequality, we have
$$
\begin{array}{l}
(a+b+c)(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) \\
\geqslant(a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a} \\
\geqslant 2 \sqrt{a b} \sqrt{a b}+2 \sqrt{b c} \sqrt{b c}+2 \sqrt{c a} \sqrt{c a} \\
=2(a b+b c+c a),
\end{array}
$$

with equality holding if and only if $c=0$ and $a=b$.
Since $a b+b c+c a=a+b+c>0$, we have
$$
\sqrt{a b}+\sqrt{b c}+\sqrt{c a} \geqslant 2 \text {. }
$$

By setting $c=0, a=b, a b+b c+c a=a+b+c$, we get $a=b=2$.
Therefore, when two of $a, b, c$ are 2 and one is 0, $\sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ achieves its minimum value of 2.