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. $2 \sqrt{2}$
Solution: A.
Assume $a \geqslant b \geqslant c$, by the AM-GM inequality, we have
$$
\begin{aligned}
& (a+b+c)(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) \\
= & (a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a}+(c \sqrt{a b}+a \sqrt{b c}+b \sqrt{c a}) \\
\geqslant & (a+b) \sqrt{a b}+(b+c) \sqrt{b c}+(c+a) \sqrt{c a} \\
\geqslant & 2 \sqrt{a b} \cdot \sqrt{a b}+2 \sqrt{b c} \cdot \sqrt{b c}+2 \sqrt{c a} \cdot \sqrt{c a}=2(a b+b c+c a)
\end{aligned}
$$

Equality holds if and only if $c=0$ and $a=b$.
Since $a b+b c+c a=a+b+c>0$, we have $\sqrt{a b}+\sqrt{b c}+\sqrt{c a} \geqslant 2$.
By setting $c=0, a=b, a b+b c+c a=a+b+c$, we get $a=b=2, c=0$.
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.