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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 $A = \sum_i a_i$. 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 $B$.

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  • $\begingroup$ Take $a= (30,15,3), b = (20,9,2)$ to see that the inequality does NOT hold. $\endgroup$ – Dr. Wolfgang Hintze Feb 16 at 8:56
  • $\begingroup$ Gah. Sorry about that. I have edited the question. $\endgroup$ – Navin K. Feb 16 at 9:06

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