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 :)

It's easy to notice that the function $f(x) =\prod_{k=0}^n |x-k|$ is a symmetric function at $x = \frac{n}{2}$ (in other words, $f(x) = f(n-x)$). So, it suffice to prove $f(x) \le \frac{(n-1)!}{2}$   in the interval $x \in [1, \frac{n}{2}]$.
For $x \in [p,p+1)$ with $p\in \{0,...,[\frac{n}{2}]-1\}$, we have
$$ f(x+1) = \prod_{0 \le k \le n}|(x+1)-k| = \frac{x+1}{|x-n|}\prod_{0 \le k \le n}|x-k| = \frac{x+1}{|x-n|} f(x)$$
As $x \in [p,p+1)$ with $p\in [0,...,([\frac{n}{2}]-1)]$, we have then $x < \frac{n-1}{2}$, so $\frac{x+1}{|x-n|} <1$ or $f(x+1) < f(x)$.
We deduce that for all $p\in \{0,...,[\frac{n}{2}]-1\}$
$$\max_{x \in [p+1,p+2)}f(x) = \max_{x \in [p,p+1)}f(x+1) < \max_{x \in [p,p+1)}f(x)$$
or
$$\max_{x \in [1,\frac{n}{2}]}f(x) = \max_{x \in [1,2)}f(x)$$
So, it suffices to prove $f(x) \le \frac{(n-1)!}{2}$   in the interval $x \in [1, 2)$.
In this interval, we have
$$f(x)  = \prod_{0 \le k \le n}|x-k| = x(x-1)(2-x) \prod_{3 \le k \le n}(k-x) = x(x-1)(4-2x) \frac{1}{2}\prod_{3 \le k \le n}(k-x)   \tag{1}$$
By AM-GM inequality,
$$x(x-1)(4-2x) \le \left(\frac{x + (x-1) + 2(2-x)}{3}  \right)^3 = 1 \tag{2}$$
and because $x \in [1, 2)$, then
$$\prod_{3 \le k \le n}(k-x) \le \prod_{3 \le k \le n}(k-1) = (n-1)! \tag{3}$$
From (2) and (3), we deduce that (1) holds true. Hence, for all $x\in [1,n-1]$
$$f(x) \le \frac{(n-1)!}{2}$$
Note: The equality can't occur because the equality in (2) occurs if $x = x-1 = 4-2x$, which is impossible.
