不妨设$$a\geqslant b\geqslant c$$  $$\left( a+b+c \right)\left( \sqrt{ab}+\sqrt{bc}+\sqrt{ca} \right)=\left( a+b \right)\sqrt{ab}+\left( b+c \right)\sqrt{bc}+\left( c+a \right)\sqrt{ca}+\left( c\sqrt{ab}+a\sqrt{bc}+b\sqrt{ca} \right)$$  $$\geqslant \left( a+b \right)\sqrt{ab}+\left( b+c \right)\sqrt{bc}+\left( c+a \right)\sqrt{ca}$$  $$\geqslant 2\sqrt{ab}\sqrt{ab}+2\sqrt{bc}\sqrt{bc}+2\sqrt{ca}\sqrt{ca}$$  $$=2\left( ab+bc+ca \right)$$  所以，$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\geqslant 2$$，当且仅当$$c=0$$，$$a=b$$时等号成立．  由$$c=0$$，$$a=b$$，解得$$a=b=2$$，$$c=0$$．  所以，$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$$的最小值为$$2$$．  故选$$\text{A}$$． 
设非负实数$$a$$，$$b$$，$$c$$满足$$ab+bc+ca=a+b+c\textgreater0$$，则$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$$的最小值为（~ ~ ）．

A: $$2$$ 
B: $$3$$ 
C: $$\sqrt{3}$$ 
D: $$2\sqrt{2}$$ 