Trying to prove that $\frac{n}{n+1}$ is a global maximum of my function I have got the function $f: \mathbb R \rightarrow \mathbb R: x \mapsto x^n(1-x) $ and I am trying to prove that for odd values of $n$, I have a global maximum at    $x = \frac{n}{n+1}$. I tried to take any $\epsilon > 0 $ and set $x$ to $x = \frac{n}{n+1} - \epsilon$ . But when I try to estimate whether or not $f(x)$ is actually smaller than $f(\frac{n}{n+1})$, I get stuck.
For example, I have $$f(\frac{n}{n+1}- \epsilon) = (\frac{n}{n+1}- \epsilon)^n-(\frac{n}{n+1}- \epsilon)^{n+1}$$
In case that $\epsilon < \frac{n}{n+1}$, while the first summand is truly smaller than $(\frac{n}{n+1})^n$, I am at the same time substracting something smaller so I cannot say for sure that it is smaller than $f(\frac{n}{n+1})$. Are there any inequalities, which could come in handy here?
Any help would be greatly appreciated!
 A: For $x\in[0,1]$, you can use the Cauchy AM-GM inequality for $n+1$ numbers:
$$x^n(1-x)=n^n\left(\frac{x}{n}\right)^n(1-x)\le n^n\left(\frac{\frac{x}{n}+\dots+\frac{x}{n}+(1-x)}{n+1}\right)^{n+1}=n^n\left(\frac{1}{n+1}\right)^{n+1}$$
Since the equality can hold (for $\frac{x}{n}=1-x$), the result obtained is indeed a maximum, and it is $f(\frac{n}{n+1})$.
A: It might not be what OP asks for, but I will leave my answer using the derivative. Maybe someone will find it helpful.

The derivative is $$f'(x)=nx^{n-1}(1-x)-x^n=x^{n-1}\left[n(1-x)-x\right]=x^{n-1}\left[n-(n+1)x\right]$$ Equating to zero one have $x_0=0\, , \, x_1=\dfrac{n}{n+1}$. Now, \begin{align*}f'(-1)&=(-1)^{n-1}\left[2n+1\right]>0 \\ f'\left(\frac{n}{2(n+1)}\right)&=\left(\frac{n}{2(n+1)}\right)^{n-1}\cdot\frac{n}{2}>0\\ f'\left(\frac{2n}{n+1}\right)&=\left(\frac{2n}{n+1}\right)^{n-1}\cdot(-n)<0\end{align*}
where I used the fact that $n$ is odd in the computation of $f'(-1)$ [if $n$ is even, then $x_0$ is a minimum point]. Based on the results above, $x_1$ is a maximum point.
