2ND EDIT: It appears that what's below was based on a misunderstanding of the problem. I thought one could permute the $A_i$ and the $B_i$ independently, whereas in fact it seems they are meant to be linked. That being the case, I suspect there is no algorithm significantly more efficient than just testing every permutation, which is to say, there is no efficient algorithm. But that's just a guess, I'm nowhere near being able to give a proof.
So it looks like you wind up with $A_1+A_2B_1+A_3B_1B_2+\cdots$ The sequence $B_1,B_1B_2,B_1B_2B_3,\dots$ is decreasing, so I think you want it to decrease as slowly as possible, so take $B_1\ge B_2\ge B_3\cdots$. Then to make the most of the $A_i$, take $A_1\ge A_2\ge A_3\cdots$.
Note that my subscripts are in reverse order; my $A_1$ is the last item chosen.
EDIT: the procedure above maximizes, the problem was to minimize, so, back to the drawing board. Also, I acted as though the base case was 1, when it's actually zero. And not only that, I thought I was reversing the order of the $A_i$, but I wasn't. So, here's what you do:
First, permute so $B_n$ is the largest of the $B_i$, because it's going to get multiplied by zero. Permute the other $B_i$ so that $B_1\le B_2\le\dots\le B_{n-1}$; that will minimize all the products of the $B_i$. Then, you want the biggest $A_i$ multiplying the smallest of the $B$-products, the smallest $A_i$ multiplying the biggest of the $B$-products, so permute the $A_i$ so that $A_1\le A_2\le A_3\le\dots\le A_n$.
See the example worked out in the comments.