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!


1 Answer 1


Using AM-GM:

$$\sqrt[n+1]{(1-x)\left(\frac{x}{n}\right)^n}\leq \frac{1-x+\frac{x}{n}+\ldots+\frac{x}{n}}{n+1}=\frac{1-x+n\cdot \frac{x}{n}}{n+1}=\frac{1}{n+1}$$

and the conclusion follows after raising to the ($n+1$)-th power.

  • 2
    $\begingroup$ Yes, we can prove AM-GM without derivative! :) +1 $\endgroup$ Feb 16, 2020 at 20:32

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