We say that a $\sigma\in S_n$ is a cool permutation if $\sigma(i)\in \{i-1,i,i+1\}$ for each $i$. For example, the following are all the cool permutations for $n=6$. (Here I write the image of $k$ in the $k$-th position)
- 1,2,3,4,5,6
- 1,2,3,4,6,5
- 2,1,3,4,5,6
- 1,3,2,4,5,6
- 1,2,4,3,5,6
- 1,2,3,5,4,6
- 1,2,4,3,5,6
- 2,1,3,4,6,5
- 2,1,4,3,5,6
- 1,3,2,5,4,6
- 1,3,2,4,6,5
- 2,1,3,5,4,6
- 2,1,4,3,6,5
I colored in the fixed points under the permutations. In this case, there is $1$ permutation that fixes everything, $5$ permutations that fix $4$ elements, $6$ permutations that fix $2$ elements and $1$ permutation that fixes $0$ elements. Now, consider the following function $f_\sigma:\mathbb{N}\to\{0,1\}$ for every $\sigma\in S_n$
$$ f_\sigma(i)= \begin{cases} 1\;\text{ if }\sigma(i)=i\\ 0\;\text{ otherwise} \end{cases} $$ And consider the polynomials
$$p_n(x_1,\dots,x_n)=\sum_{\sigma\text{ is cool}}\prod_{i=1}^nx_i^{f_\sigma(i)}$$ For instance, using the list above, one can compute $$p_6(x_1,\cdots,x_6)=x_1\cdots x_6+x_1x_2x_3x_4+x_3x_4x_5x_6+\cdots+x_3x_6+1$$ I want to prove that for every $x_1,\cdots,x_n$ integers greater than $1$, one has $$p_n(x_1,\cdots,x_n)-p_{n-2}(x_2,\cdots,x_{n-1})>\max\left\{p_{n-1}(x_1,\cdots,x_{n-1}),p_{n-1}(x_2,\cdots,x_n)\right\}$$ This is of course true when all $x_i$ are large enough, because the degree of the LHS is greater than the one of the RHS, I want to show that $x_i>1$ for every $i$ is "large enough". I've proved this is the case for $n\in\{1,\cdots,5\}$, but I could go forever and not get anywhere.
Is this hard? This arose while I was trying to prove something about periodic continued fractions; if I'm not lucky with this approach maybe I'll just ask the original question