How do we prove that $(x-1)!leq{(frac{x}{2})^{x-1}}$?
0
$begingroup$
I plotted the graph of $y=(x-1)!$ and $y=(frac{x}{2})^{x-1}$ and found that the latter is always greater than the former for $x>2$ . Also, equality holds at $x=2$ . Would someone please help me to prove it mathematically?
inequality
share | cite | improve this question
edited Jan 6 at 9:37
Shashwat1337
asked Jan 5 at 18:26
Shashwat1337 Shashwat1337
95 9
$endgroup$
...