proving an inequality based on double products and binomial 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)?
 A: 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}
$
