TITLE: Is $ \left(\frac{1}{2}\right)^n < \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$ true for $n \ge 3$?
QUESTION [8 upvotes]: I have reason* to believe that the following inequality holds:
$$ \left(\frac{1}{2}\right)^n < \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}$$
for every $n\ge3$. I am having hard time proving it... (induction did not work).
(Comment: For $n=2$ there is an equality).

*My reasoning is very convoluted and is related to an entirely different problem, I am sure there is a straightforward argument.

REPLY [5 votes]: If $f(x)=x(1-x)^{n-1}$ then $$f'(x)=(1-x)^{n-1}-x(n-1)(1-x)^{n-2}=(1-x)^{n-2}(1-nx)$$
So $f$ has a critical point where $x=\frac{1}{n}$. Since $f(0)=f(1)=1$ and this is the only critical point in $(0,1)$, and $f(x)>0$ in $(0,1)$, you have that $f(x)$ is maximized when $x=\frac{1}{n}.$ In particular, then:
$$f\left(\frac12\right)<f\left(\frac 1n\right)$$
This is actually true for any real $n>1$ except $n=2$ when $\frac{1}{2}=\frac{1}{n}$.
If you have a weighted coin with probability $p$ of coming up heads, then $np(1-p)^{n-1}$ is the probability than in $n$ tosses, you'd get exactly one heads. Then what the above argument says is that, to maximize the odds of getting one heads in $n$ tosses, you'd want $p=\frac{1}{n}$, which is somewhat intuitive...