Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$

Question

Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$.

My try: $f'(x)=nx^{n-1}-(n+1)x^n=x^{n-1}(n-(n+1)x)=0\Rightarrow x=\frac{n}{n+1} $, hence $\max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{1}{n+1})^n\cdot\frac{1}{n+1}$. I know $(1-\frac{1}{n+1})^n\to e^{-1}$. But I don't know how to prove $(1-\frac{1}{n+1})^n\lt \frac{1}{e}$.

Answer 1

Hint: Prove that $g(x)=x^x$ is increasing, then with $n<n+1$ $$(1-\frac{1}{n+1})^n\cdot\frac{1}{n+1}=\dfrac{n^n}{(n+1)^{n+1}}<1$$ Edit: You want to prove $$\left(1+\dfrac{1}{n}\right)^{n+1}>e$$ by mean value theorem for $\ln x$ $$\ln(n+1)-\ln(n)=\dfrac{1}{\xi}>\dfrac{1}{n+1}$$ where $n<\xi<n+1$.

Answer 2

\begin{equation} \max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{n}{n+1})^n\cdot\frac{n}{n+1} = \frac{n}{(n+1)^{n+1}} \tag{1} \end{equation} Now \begin{equation} \frac{n}{(n+1)^{n+1}} - \frac{1}{ne} = \frac{1}{n}( \frac{n^2}{(n+1)^{n+1}} - \frac{1}{e}) = \frac{1}{n}( \frac{en^2 - (n+1)^{n+1}}{e(n+1)^{n+1}}) \tag{2} \end{equation} Sign depends on $en^2 - (n+1)^{n+1}$. It is easy to see that for $n=1$, $en^2 - (n+1)^{n+1} < 0$. We can say that for $n \geq 2$, \begin{equation} \begin{split} en^2 - (n+1)^{n+1} &< e(n+1)^2 - (n+1)^{n+1} \\ &= (n+1)^2(e - (n+1)^{n-1}) <0 \end{split} \end{equation} So for all $n \geq 1$, we have that $en^2 - (n+1)^{n+1} < 0$. Going back to $(2)$, we get $\frac{n}{(n+1)^{n+1}} - \frac{1}{ne} < 0$, which using $(1)$, gives \begin{equation} \max_{x\in(0,1)} f(x) < \frac{1}{ne} \end{equation}