# A curious inequality concerning binomial coefficients

Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.

Let $$a_1,a_2,\ldots,a_k$$ be non-negative integers such that $$\sum_i a_i = A$$. Then, for any non-negative integer $$B \le A$$: $$\sum_{(b_1,\ldots,b_k): \sum_i b_i = B} \prod_i \frac{\binom{a_i}{b_i}}{\binom{A-a_i}{B-b_i}} \ge {\binom{A}{B}}^{2-k}.$$ The sum on the left is over all tuples $$(b_1,b_2,\ldots,b_k)$$ of non-negative integers, with $$b_i \le a_i$$ for all $$i$$, whose sum is equal to $$B$$.

By Cauchy–Bunyakovsky–Schwarz inequality we have $$\left(\sum \prod_i \frac{\binom{a_i}{b_i}}{\binom{A-a_i}{B-b_i}}\right)\left(\sum\prod_i \binom{a_i}{b_i}\binom{A-a_i}{B-b_i}\right)\geqslant \left(\sum \prod_i\binom{a_i}{b_i}\right)^2=\binom{A}{B}^2 .$$
Thus it suffices to prove that $$\sum\prod_i \binom{a_i}{b_i}\binom{A-a_i}{B-b_i}\leqslant \binom{A}B^k.$$ But RHS is just the sum of the same guys without the restriction $$\sum b_i=B$$.