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My previous question was solved in a very elegant way, hopefully this (seemingly more complicated) case is also easy for experts.

I need the inequality $\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\bigg(\sum\limits_{\substack{(k_1,\dots,k_r),\ k_i\ge0\\\sum_i k_i=n-j}}\prod_{i=1}^r(p_i-1)^{k_i}\bigg)-(-1)^n\ge A_{n,r}\Big(\prod^r_{i=1}p_i\Big)\sum\limits_{\substack{k_1,\dots, k_r\geq 0\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{1}{k_i+1}$

Here $n\ge3$ and $A_{n,r}:=\frac{\binom{n+r−1}{n}(n+r)!}{r!S_{n+r,r}}$, with $S_{n+r,r}$- Stirling numbers of the second kind. Some properties of $A_{n,r}$ are given here. The numbers $n,r,p_i$ are positive integers.

I tried various naive approaches (some inductions or to check that some things cancel out). With no success. Though I can prove the inequality for $r=1$ or for $n=3$. Already the case $n=4$ gets too messy. (And of course I checked numerically for a variety of cases.) Even a simpler version would be quite helpful:

$\sum\limits_{\substack{\{k_i\ge0\}\\\sum_i k_i=n}}\prod_{i=1}^r(p_i-1)^{k_i}\ge A_{n,r}\sum\limits_{\substack{\{k_i\geq 0\}\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{1}{k_i+1}$

Any suggestions?

upd. We succeeded to prove this inequality, though in a terribly complicated way. Still I would like to see some short proof.
show/hide this revision's text 2 some typos corrected

My previous question was solved in a very elegant way, hopefully this (seemingly more complicated) case is also easy for experts.

I need the inequality $\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\bigg(\sum\limits_{\substack{(k_1,\dots,k_r),\ k_i\ge0\\\sum_i k_i=n-j}}\prod_{i=1}^r(p_i-1)^{k_i}\bigg)-(-1)^n\ge A_{n,r}\Big(\prod^r_{i=1}p_i\Big)\sum\limits_{\substack{k_1,\dots, k_r\geq 0\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{1}{k_i+1}$

Here $n\ge3$ and $A_{n,r}:=\frac{\binom{n+r−1}{n}(n+r)!}{r!S_{n+r,r}}$, with $S_{n+r,r}$- Stirling numbers of the second kind. Some properties of $A_{n,r}$ are given here. The numbers $n,r,p_i$ are positive integers.

I tried various naive approaches (some inductions or to check that some things cancel out). With no success. Though I can prove the inequality for $r=1$, $r=2$, r=1$ or for $r=3$. n=3$. Already the case $r=4$ n=4$ gets too messy. (And of course I checked numerically for a variety of cases.) Even a simpler version would be quite helpful:

$\sum\limits_{\substack{\{k_i\ge0\}\\\sum_i k_i=n}}\prod_{i=1}^r\binom{p_i-1}{k_i}k_i!\ge k_i=n}}\prod_{i=1}^r(p_i-1)^{k_i}\ge A_{n,r}\sum\limits_{\substack{\{k_i\geq 0\}\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{1}{k_i+1}$

I am trying to write the generating functions for the later inequality. For the right hand side I have: $\sum_{n\ge0}t^{n+r}\binom{n+r-1}{n}\sum_{\{p_i\ge1\}}\sum\limits_{\substack{\{k_i\geq 0\}\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{x^{p_i-1}_i}{k_i+1}=\frac{t\partial^{r-1}_t}{(r-1)!}\prod^r_{i=1}\Big(-\frac{\ln(1-\frac{xt_i}{1-x_i})}{x_i}\Big)$.

For the left hand side: $S_{n+r,r}r!\frac{\sum_{\substack{(k_1,\dots,k_r),\ k_i\ge0\\\sum_i k_i=n}}\prod_{j=1}^r\binom{p_i-1}{k_i}k_i!}{(n+r)!}=\frac{r!}{\prod^r_{i=1}x_i}\sum_{n\ge0}\frac{S_{n+r,r}r!}{(n+r)!} \sum\limits_{\substack{\{k_i\ge0\}\\\sum_i k_i=n}}\prod_{j=1}^r\Big(\frac{tx_i}{1-x_i}\Big)^{k_i+1}k_i!$

Is this later expression the Taylor expansion of some nice function?

Any other suggestions?
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A combinatorial bound involving Stirling numbers of the second type

My previous question was solved in a very elegant way, hopefully this (seemingly more complicated) case is also easy for experts.

I need the inequality $\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\bigg(\sum\limits_{\substack{(k_1,\dots,k_r),\ k_i\ge0\\\sum_i k_i=n-j}}\prod_{i=1}^r(p_i-1)^{k_i}\bigg)-(-1)^n\ge A_{n,r}\Big(\prod^r_{i=1}p_i\Big)\sum\limits_{\substack{k_1,\dots, k_r\geq 0\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{1}{k_i+1}$

Here $A_{n,r}:=\frac{\binom{n+r−1}{n}(n+r)!}{r!S_{n+r,r}}$, with $S_{n+r,r}$- Stirling numbers of the second kind. Some properties of $A_{n,r}$ are given here. The numbers $n,r,p_i$ are positive integers.

I tried various naive approaches (some inductions or to check that some things cancel out). With no success. Though I can prove the inequality for $r=1$, $r=2$, $r=3$. Already the case $r=4$ gets too messy. (And of course I checked numerically for a variety of cases.) Even a simpler version would be quite helpful:

$\sum\limits_{\substack{\{k_i\ge0\}\\\sum_i k_i=n}}\prod_{i=1}^r\binom{p_i-1}{k_i}k_i!\ge A_{n,r}\sum\limits_{\substack{\{k_i\geq 0\}\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{1}{k_i+1}$

I am trying to write the generating functions for the later inequality. For the right hand side I have: $\sum_{n\ge0}t^{n+r}\binom{n+r-1}{n}\sum_{\{p_i\ge1\}}\sum\limits_{\substack{\{k_i\geq 0\}\\\sum_i k_i=n}}\prod^r_{i=1}\binom{p_i-1}{k_i}\frac{x^{p_i-1}_i}{k_i+1}=\frac{t\partial^{r-1}_t}{(r-1)!}\prod^r_{i=1}\Big(-\frac{\ln(1-\frac{xt_i}{1-x_i})}{x_i}\Big)$.

For the left hand side: $S_{n+r,r}r!\frac{\sum_{\substack{(k_1,\dots,k_r),\ k_i\ge0\\\sum_i k_i=n}}\prod_{j=1}^r\binom{p_i-1}{k_i}k_i!}{(n+r)!}=\frac{r!}{\prod^r_{i=1}x_i}\sum_{n\ge0}\frac{S_{n+r,r}r!}{(n+r)!} \sum\limits_{\substack{\{k_i\ge0\}\\\sum_i k_i=n}}\prod_{j=1}^r\Big(\frac{tx_i}{1-x_i}\Big)^{k_i+1}k_i!$

Is this later expression the Taylor expansion of some nice function?

Any other suggestions?