
> 
> **Question.** Let $a, b, c\geq0$ be integers. Does this inequality hold?
>  $$\binom{(a+b+2)(a+c+3)+1}{c+3}\geq\binom{a+c+3}{c+3}(a+b+3)^{c+3}.$$
> 
> 
> 


This inequality happens to appear in some intermediate step involved in [this work](https://www.math.temple.edu/~tewodros/Catalan_Convexity.pdf). Thank you for any suggestions.



inequality with binomials by breaking ups
I think we can do this combinatorially, but it's messier than I thought originally and with my current crude version I can't treat the full range. After relabeling, the inequality becomes
$$
{mn+1 \choose k} \ge {n\choose k} (m+1)^k ,
$$
for $n\ge k\ge 3$, $m\ge n-k+2$. So $m\ge 2$, and I can handle $m\ge 4$, which I want to assume from now on.


On the LHS, I just choose $k$ chips out of $mn+1$. On the RHS, I'll put groups of $m+1$ chips each into $n$ boxes, with overlap of one chip from one box to the next. In other words, both boxes $1$ and $2$ have a copy of chip $m+1$ in them, and chip $2m+1$ is in boxes $2$ and $3$ etc.


Now I pick $k$ boxes, and then select one of the $m+1$ chips out of each box. I have RHS many ways of doing this. Whenever a chip was chosen twice, I replace this with an otherwise impossible choice of two chips from the smaller box (for example, the second and third chips from that box).


I have produced some, but not all selections of $k$ chips. There is the additional issue that I can get certain configurations in more than one way: for example, chips $m+1$, $2m+1$ I can get either from boxes $1,2$ or from boxes $2,3$. This double counting is of course harmless here since there are way more configurations I can't get at all. More formally, I can again replace the first alternative with an otherwise impossible choice of two chips from the first box (say chips $3,4$ from that box).


