Factorial inequality.

Question

I have the following factorial function

$(m-k-1)!(k-1)!$ for $m \in \mathbb{N}$, $m \geq 2$ and $k \in \{1,\cdots,m-1\}$.

I'm trying to find the value of $k$ for which the above expression attains minimum, or perhaps if there exists a function $f(m)$ such that

$f(m)! \leq (m-k-1)!(k-1)!$ for all $k \in \{1,\cdots,m-1\}$?

Plugging in certain values, it seems that the minimum is achieved for $k \approx m/2$, assuming $m$ is even. However, I have no idea about how to get the analytic expression.

Any help would be much appreciated. Thanks.

Answer 1

You are looking for the lowest $k$ s.t. $(m-k-2)!k! \ge (m-k-1)!(k-1)!$

So we look for $k \ge m-k-1 \implies 2k \ge m-1$

Solving for the minimum $k_{min} =$ largest integer greater than or equal to $\dfrac{m-1}{2}$.

For odd $m$, this is $\dfrac{m-1}{2}$ and for even $m$, this is $\dfrac{m}{2}$

Answer 2

This method is not original by me.

Let $f(k) =(m-k-1)!(k-1)! $.

Then $r(k) =\dfrac{f(k+1)}{f(k)} =\dfrac{(m-(k+1)-1)!((k+1)-1)!}{(m-k-1)!(k-1)!} =\dfrac{(m-k-2)!k!}{(m-k-1)!(k-1)!} =\dfrac{k}{m-k-1} $.

If $k < m-k-1$ (i.e., $k < \frac{m-1}{2}$), $r(k) < 1$ so $f(k)$ is decreasing.

If $k > m-k-1$ (i.e., $k > \frac{m-1}{2}$), $r(k) > 1$ so $f(k)$ is increasing.

If $m$ is odd, then the minimum value is at $k=\frac{m-1}{2}$.

If $m$ is even, $r(\frac{m}{2}) =\frac{m/2}{m-m/2-1} =\frac{m/2}{m/2-1} >1 $ and $r(\frac{m}{2}-1) =\frac{m/2-1}{m-(m/2-1)-1} =\frac{(m-2)/2}{(m-2)/2} =1 $.

Therefore, if $m$ is even, the minimum value of $f(k)$ is at both $k=\frac{m}{2}$ and $k=\frac{m}{2}-1$.