prove $\prod_{i=0}^{n}x-\frac{i}{n}\le\frac{n!}{4n^{n+1}}$ for any $x\in[0,1]$ and $n>2$ prove $\prod_{i=0}^{n}x-\frac{i}{n}\le\frac{n!}{4n^{n+1}}$ for any $x\in[0,1]$ and natural $n>2$.
I tried using AM-GM, but found nothing interesting.
I was looking also for a Taylor series. no luck
 A: Let $(\sigma_0,\ldots,\sigma_n)$ be a permutation of $(0,\ldots,n)$ such that $$|x-\tfrac{\sigma_0}{n}| \le |x-\tfrac{\sigma_1}{n}| \le \cdots \le |x-\tfrac{\sigma_{n-1}}{n}| \le |x-\tfrac{\sigma_n}{n}|.$$ In other words, of the points $\{\tfrac{0}{n},\tfrac{1}{n},\ldots,\tfrac{n-1}{n},\tfrac{n}{n}\}$, $\tfrac{\sigma_0}{n}$ is the closest to $x$, $\tfrac{\sigma_1}{n}$ is the 2nd closest to $x$, ..., and $\tfrac{\sigma_n}{n}$ is the farthest from $x$, with ties broken arbitrarily.
If $x = \tfrac{k}{n}$ for some $k = 0,\ldots,n$, then $\prod_{i = 0}^{n}|x-\tfrac{i}{n}| = 0 \le \tfrac{n!}{4n^{n+1}}$ holds trivially.
Now, suppose $x \in (\tfrac{k-1}{n},\tfrac{k}{n})$ for some $k = 1,\ldots,n$. Then $\{\sigma_0,\sigma_1\} = \{k-1,k\}$, and thus, $$\left|x-\tfrac{\sigma_0}{n}\right| \cdot \left|x-\tfrac{\sigma_1}{n}\right| = (x-\tfrac{k-1}{n})(\tfrac{k}{n}-x) \le \left(\dfrac{(x-\tfrac{k-1}{n})+(\tfrac{k}{n}-x)}{2}\right)^2 = \dfrac{1}{4n^2}.$$
Also, for $i = 2,\ldots,n$, we have $|x-\tfrac{\sigma_i}{n}| \le \tfrac{i}{n}$ since at least $i+1$ of the points $\{\tfrac{0}{n},\tfrac{1}{n},\ldots,\tfrac{n-1}{n},\tfrac{n}{n}\}$ are within $\tfrac{i}{n}$ of the point $x$. Hence, $$\prod_{i = 0}^{n}|x-\tfrac{i}{n}| = \prod_{i = 0}^{n}|x-\tfrac{\sigma_i}{n}| = \left|x-\tfrac{\sigma_0}{n}\right| \cdot \left|x-\tfrac{\sigma_1}{n}\right| \cdot \prod_{i = 2}^{n}\left|x-\tfrac{\sigma_i}{n}\right| \le \dfrac{1}{4n^2}\prod_{i = 2}^{n}\dfrac{i}{n} = \dfrac{n!}{4n^{n+1}}.$$
A: Let $f(x)=\prod_{i=0}^n\left|x-\frac{i}{n}\right|$. Since $f$ is continuous on the compact interval $[0,1]$, it attains its max at some $y\in[0,1]$. As $f(x)=f(1-x)$, we may assume WLOG that $y\in[0,1/2]$.
If $y>1/n$, then
$$f\left(y-\frac{1}{n}\right)=\left(1-\frac{1}{n}-y\right)\cdot\prod_{i=0}^{n-1}\left|y-\frac{1}{n}-\frac{i}{n}\right| > y\cdot\prod_{i=1}^n\left|y-\frac{i}{n}\right|=f(y).$$
So we must have $0\leq y\leq\frac{1}{n}$. Then by AM-GM, $y\left(\frac{1}{n}-y\right)\leq\frac{1}{4n^2}$, so
$$f(y)=y\left(\frac{1}{n}-y\right)\cdot\prod_{i=2}\left(\frac{i}{n}-y\right)\leq\frac{1}{4n^2}\cdot\prod_{i=2}^n\frac{i}{n}=\frac{n!}{4n^{n+1}}.$$
