Inequalities between sums of products of certain binomial coefficients

Question

I am a PhD student. During my researches, I often have to deal with inequalities involving sums of binomial coefficients, where the sums are indexed by some set of integer compositions. For example, let $l,k$ be positive integers such that $l \leq k$ and either $l$ is odd or both $l$ and $k$ are even. One of my results can be proven if I show that for any $i \leq k \lfloor \frac{l}{2}\rfloor - 1$:

$$ \displaystyle \sum_{\substack{\alpha \models i+k} \\ \ell(\alpha) = k \\ \alpha_a \leq \lfloor \frac{l}{2} \rfloor + 1} \prod_{a=1}^k \binom{l}{2(\alpha_a - 1)} \leq \displaystyle \sum_{\substack{\alpha \models i+l} \\ \ell(\alpha) = l \\ \alpha_a \leq \lfloor \frac{k}{2} \rfloor + 1} \prod_{a=1}^l \binom{k}{2(\alpha_a - 1)}, $$

where the notations under the left sum means that we take the sum over all $\alpha = (\alpha_1,...,\alpha_k) \subseteq \{1,...,\lfloor \frac{l}{2} \rfloor + 1\}^k$ such that $\alpha_1 + ... + \alpha_k = k+i$, and analogously for the other.

My question here is not about solving this problem in particular (although if the solution is easy feel free to give it).

My question is rather: how do you work with such sums? I am not used to these kinds of formulas, and while it is manageable when $l$ and $k$ are small, they become hard very rapidly. So, is there a trick to understand these kind of sums? Is there some already known results? Is there formulas? Combinatorial arguments?

Answer 1

Take the LHS: $$\sum_{\substack{\alpha \models i+k} \\ \ell(\alpha) = k \\ \alpha_a \leq \lfloor \frac{l}{2} \rfloor} \prod_{a=1}^k \binom{l}{2(\alpha_a - 1)}$$ Firstly, we can simplifying by rolling the subtraction of one into the definition of $\alpha$: $$\sum_{\substack{\sum \alpha_a = i \\ \ell(\alpha) = k \\ 0 \leq \alpha_a < \lfloor \frac{l}{2} \rfloor}} \prod_{a=1}^k \binom{l}{2\alpha_a}$$ The constraint on the sum of the $\alpha_a$ can be done with coefficient extraction: $$\begin{eqnarray*} & [x^{2i}] \sum_{\substack{\ell(\alpha) = k \\ 0 \leq \alpha_a < \lfloor \frac{l}{2} \rfloor}} \prod_{a=1}^k \binom{l}{2\alpha_a} x^{2\alpha_a} \\ =& [x^{2i}] \left( \sum_{j=0}^{\lfloor \frac{l}{2} \rfloor - 1} \binom{l}{2\alpha_a} x^{2\alpha_a} \right)^k \\ =& [x^{2i}] \left( \frac{ (1+x)^l + (1-x)^l }2 - \binom{l}{l - l \bmod 2}x^{l - l \bmod 2} \right)^k \end{eqnarray*}$$

One (untested) idea to show that the coefficients of the polynomial corresponding to the LHS don't exceed those of the RHS would be to take the $2i^{\textrm{th}}$ derivative and evaluate it at zero.