# Proof of $(1-x)x^n \leq \frac{n^n}{\left(n+1\right)^{n+1}}$ without use of derivatives

If $$x \in \left[0,1\right]$$ and $$n \in \mathbb{Z}^+$$, is it possible to show $$(1-x)x^n \leq \frac{n^n}{\left(n+1\right)^{n+1}}$$ without use of derivatives?

With derivative it's smooth:

Let $$f(x)=(1-x)x^n.$$

Thus, $$f'(x)=nx^{n-1}-(n+1)x^n=(n+1)x^{n-1}\left(\frac{n}{n+1}-x\right),$$ which gives $$x_{\max}=\dfrac{n}{n+1}$$ and we are done!

$$\sqrt[n+1]{(1-x)\left(\frac{x}{n}\right)^n}\leq \frac{1-x+\frac{x}{n}+\ldots+\frac{x}{n}}{n+1}=\frac{1-x+n\cdot \frac{x}{n}}{n+1}=\frac{1}{n+1}$$
and the conclusion follows after raising to the ($$n+1$$)-th power.