Let $\epsilon \in \mathbb{R}$, $0 \le \epsilon \le 1$. Let $n$ be a positive integer. Let $k$ be a non-negative integer, $0 \leq k \leq n+1$.
Is it true that $|(1 - k\epsilon)(2 - k\epsilon)...(n - k\epsilon)| \leq n!$
I tried some numbers and it seems to work but I am not sure if it's true in general. Any hints would be appreciated.
On
Let's replace $n+1$ by $n$ and $k\epsilon$ by $x$. Then the question is:
For an integer $n>1$, and $x\in\mathbb{R}$, let $f_n(x)=\prod_{k=1}^{n-1}|k-x|$.
Show that $f_n(x)\leqslant f_n(0)$ for $0\leqslant x\leqslant n$.
We may assume that $x$ is not an integer. Since $f_n(x)=f_n(n-x)$, we may also assume that $x\leqslant n/2$. Now, if $x>1$ then $f_n(x)/f_n(x-1)=(x-1)/(n-x)<x/(n-x)\leqslant 1$. Thus, by induction (i.e. after repeated decrease of $x$ by $1$), we may assume that $0<x<1$. Finally, $f_n(x)$ is clearly decreasing on this interval, and we're done.
On
Consider the integer lattice consisting of the points $1,\dots,n$. We shift this through the reals until the point $1$ is on the point $-n$.
Each point in the lattice that is positive is less than or equal to the nearest integer above or equal to it.
Each point in the lattice that is negative, add $n$ to it. Then they are less than or equal to the remaining integers from the original lattice.
We have $0 \le k\epsilon < n+1$.
For $k \epsilon \ge 1$, we take $m = \lfloor k \epsilon \rfloor$. For $0 \le k \epsilon < 1$, we take $m = 1$.
Either way we have $|m - k \epsilon| \le 1$ for some $1 \le m \le n$.
Now we have: $$|1-k\epsilon|\le |m-k\epsilon| + |1-m| \le m$$ $$:$$ $$|(m-2)- k\epsilon| \le |m- k\epsilon|+|-2| \le 3$$ $$|(m-1)- k\epsilon| \le |m- k\epsilon|+|-1| \le 2$$ $$|m- k\epsilon| \le 1$$ $$|(m+1)- k\epsilon| \le |m- k\epsilon|+|1| \le 2\le m+1$$ $$|(m+2)- k\epsilon| \le |m- k\epsilon|+|2| \le 3 \le m+2$$ $$:$$ $$|n - k\epsilon| \le |m-k\epsilon| + |n-m| \le n-m+1\le n$$
Hence we see that $|(1-k\epsilon)(2-k\epsilon)\dots(n-k\epsilon)| \le m!(n-m+1)! \le n!$ by term-by-term comparison.