Smart way or procedure to upperbound product of two summands?

142 Views Asked by At

Let $a_1,\dots,a_n$ and $b_1,\dots,b_n$ be positive real numbers. I would like to handle products of this form: $$\prod_{i=1}^{n}\left(a_i+b_i\right).$$

It looks like there are $2^n$ summands where each term has $n$ factors and $a_i$ and $b_i$ appear in them. Is there a way to handle/write down this product?

Thanks for your help!

1

There are 1 best solutions below

0
On BEST ANSWER

Let $[n]:=\{1,2,\dots,n\}$. Then we can expand the finite product as $$\prod_{i=1}^{n}(a_i+b_i) = \sum_{A\;\subseteq\;[n]}\Big(\prod_{j\;\in \;A}a_j\prod_{k\;\in\;[n]\setminus A}b_k\Big).$$ You may want to list the subsets $A$ of $[n]$ in some particular order. Using permutations of $[n]$ gives similar but different orders.

We are given that $\;a_1,\dots,a_n\;$ and $\;b_1,\dots,b_n\;$ are positive real numbers. Without more specfic information about these numbers, it is not possible to get an upper found on the product of the pairwise sums. One obvious example would be that if $\;a_i\le A,\;b_i\le B\quad \forall i\in[n]\;$ then the product is bounded by $\;(A+B)^n.$