proving an inequality based on double products and binomial

Question

Iv been trying to prove the following inequality: let $a_1,\ldots,a_n$ be a non-increasing sequence, i.e., $a_1\geq a_2\geq \cdots \geq a_n\geq 0$ such that $\sum_i a_i=m$. Then, prove that $$\prod_{i=1}^n\prod_{j=i+1}^{n} \big(1+\frac{a_i-a_j}{j-i}\big)\leq \binom{n+m}{m}.$$ I tried proving it by induction on the number of non-zero entries in $a$, since I see how to prove the inequality in the simple case when $a_1=m$ and the remaining $a_i$s are zero. Proving it more generally seems challenging. Does anyone see how to prove such an inequality (or what is a good starting point to show this)?

Answer 1

This is false in general.

Want to show $p(a, n) =\prod_{i=1}^n\prod_{j=i+1}^{n} \big(1+\frac{a_i-a_j}{j-i}\big) \leq \binom{n+m}{n}. $

I decided to do the special case $a_i=n-i$. $m =\sum_{i=1}^n (n-i) =\sum_{i=0}^{n-1} i =n(n-1)/2 $.

For this case

$\begin{array}\\ p(a, n) &=\prod_{i=1}^n\prod_{j=i+1}^{n} \big(1+\frac{a_i-a_j}{j-i}\big)\\ &=\prod_{i=1}^n\prod_{j=i+1}^{n} \big(1+\frac{(n-i)-(n-j)}{j-i}\big)\\ &=\prod_{i=1}^n\prod_{j=i+1}^{n} \big(1+\frac{j-i}{j-i}\big)\\ &=\prod_{i=1}^n\prod_{j=i+1}^{n} 2\\ &=\prod_{i=1}^n2^{n-i}\\ &=2^{\sum_{i=1}^n(n-i)}\\ &=2^{\sum_{i=0}^{n-1} i}\\ &=2^{n(n-1)/2}\\ \end{array} $

So we want

$\begin{array}\\ 2^{n(n-1)/2} &\le \binom{n+n(n-1)/2}{n}\\ &=\dfrac{(n+n(n-1)/2)!}{n!(n(n-1)/2)!}\\ &=\dfrac{\prod_{k=1}^{n+n(n-1)/2}k}{\prod_{k=1}^{n}k\prod_{k=1}^{n(n-1)/2}k}\\ &=\dfrac{\prod_{k=n+1}^{n+n(n-1)/2}k}{\prod_{k=1}^{n(n-1)/2}k}\\ &=\dfrac{\prod_{k=1}^{n(n-1)/2}(k+n)}{\prod_{k=1}^{n(n-1)/2}k}\\ &=\prod_{k=1}^{n(n-1)/2}\dfrac{k+n}{k}\\ &=\prod_{k=1}^{n(n-1)/2}(1+\frac{n}{k})\\ \end{array} $

For $1 \le n \le 6$ this is true, but, according to Wolfram Alpha, for $n=7$,

$\begin{array}\\ 2^{n(n-1)/2} &=2^{21}\\ &=2097152\\ \text{and}\\ \binom{n+n(n-1)/2}{n} &=\binom{28}{7}\\ &=\prod_{k=1}^{n(n-1)/2}(1+\frac{n}{k})\\ &=\prod_{k=1}^{21}(1+\frac{7}{k})\\ &=1108040\\ &\lt 2^{21}\\ \end{array} $