Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$ for $1 \le x \le n-1$. 
Let $n \ge 3$, $x \in \Bbb R$ such that $1 \le x \le n-1$. Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$.

For $n=3$, $1 \le x \le 2$ we want to show $|x(x-1)(x-2)(x-3)|=x(x-1)(2-x)(3-x) \le 1$.
We see that some large bounds are easy to guess:
$$
x(x-1)(2-x)(3-x) \le 2 \times 1 \times 1 \times 2=4
$$
but not precise enough. Another try:
$$
x(x-1)(2-x)(3-x) \le x(3-x) \le 9/4
$$
the last inequality is from the study of $f(x)=x(3-x)$.
Same problem.
For $n=3$ and $1 \le x \le 2$ let $0 \le y=x-1 \le 1$.
We get using AM-GM
\begin{align}
|x(x-1)(x-2)(x-3)| &= |(y+1)(y)(y-1)(y-2)|\\
&=y(1-y^2)(2-y)\\
&\le (\frac{3-y^2}{3})^3\\
&\le  1
\end{align}
For the general case I tried to use the same approach:
let $1 \le x \le n-1$ ie. $0 \le y = (x-1)/(n-2) \le 1$. We have
\begin{align}
\prod_{k=0}^n |x-k|
&=
\prod_{k=0}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l |(n-1)y +1-k| \prod_{k=l+1}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l ((n-1)y +1-k) \prod_{k=l+1}^n (k-1-(n-1)y)\\
\end{align}
where $l = \lfloor (n-1)y+1\rfloor$. Applying AM-GM doesn't really help here. Suppose $y<1$ ie. $l<n$, we see that
$$
\prod_{k=0}^l ((n-1)y +1-k) \le \prod_{k=0}^l (n-k)=\frac{n!}{(n-l-1)!}
$$
and
$$
\prod_{k=l+1}^n (k-1-(n-1)y) \le \prod_{k=l}^{n-1} (k) = \frac{(n-1)!}{(l-1)!}
$$
So
$$
\prod_{k=0}^n |x-k|\le
\frac{(n-1)!}{(l-1)!}\frac{n!}{(n-l-1)!}
$$
The bound is too big...
I also tried this approach: let $f(x)= \prod_{k=0}^n (x-k)$, $f'(x)=f(x)(\sum_{k=0}^n \frac{1}{x-k})$.
I'm trying to find informations about $x$ such that $f'(x)=0$. Let $x \in \Bbb R - [|0,n|]$. So $\sum_{k=0}^n \frac{1}{x-k}=0$. But I'm stuck here.
My questions are:

*

*Does someone have a hint or a proof?

*Can someone give me some advice on how to handle such problems during an exam?

Thanks :)

Consider the function $$f(x)=\prod_{k=0}^n |x-k|, \qquad x\in[1, n-1].$$ This is just a absolute value of a polynomial on a bounded closed interval, and therefore by extreme value theorem it must have a maximum  and a minimum at least once. Clearly the minimum value of $f$ is zero and achieve at each integer $1, 2, \cdots, n-1.$ Also, due to the absolute value $f(x)=f(n-x)$ and hence the graph of $f$ is symmetric about the vertical line $x=n/2.$
When $k\lt x\lt k+1$ for some $k\in\mathbb{N}$ in its domain, then by AM-GM inequality we have $$|x-k||x-(k+1)|\le\dfrac{1}{4}\left((x-k)+((k+1)-x)\right)^2=\dfrac14.$$ Or you can consider the minimum value of convex quadratic function $y=(x-k)(x-k-1)$ to get the same bound. Furthermore, we can find bounds for remaining factors by looking at distances. Hence
$$1\lt|x-(k-1)|\lt2\lt|x-(k-2)|\lt3\lt\cdots\lt k \lt|x|\lt k+1$$ and
$$1\lt|x-(k+2)|\lt2\lt|x-(k+3)|\lt3\lt\cdots\lt n-k-1 \lt|x-n|\lt n-k.$$
We can use these information to compute an upper bound for $f$ at any interval as follows.

*

*$k=1$: $\quad f(x)\lt2.\dfrac{1}{4}. 2.3.\cdots(n-1)=\dfrac{1}{2}(n-1)!$


*$k=2$: $\quad f(x)\lt3.2. \dfrac{1}{4}.2. \cdots(n-2)=\dfrac{3}{2}(n-2)!$


*$k=3$: $\quad f(x)\lt4.3.2. \dfrac{1}{4}.2. \cdots(n-3)=6(n-3)!.$
Similarly

*

*$k=r$: $\quad f(x)\lt\dfrac{1}{4}(r+1)!(n-r)!$


*$k=\lfloor n/2 \rfloor$: $\quad f(x)\lt\dfrac{1}{4}(\lfloor n/2 \rfloor+1)!(n-\lfloor n/2 \rfloor)!.$
It is pretty easy to see that the sequence $g(r)=\dfrac{1}{4}(r+1)!(n-r)!$ on $r=1, 2, \cdots, \lfloor n/2 \rfloor$ is decreasing since $$\dfrac{g(r+1)}{g(r)}=\dfrac{r+2}{n-r}\le1.$$
Hence the maximum strict upper bound of $f$ occurs on $(1,2)$ and $(n-2, n-1)$ and it is exactly $$g(1)=\dfrac{1}{2}(n-1)!.$$
