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 $B$.



A curious inequality concerning binomial coefficients
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 $\sum b\_i=B$.


