Prove that $n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!}$

Question

Let $n$ be a positive integer. I conjectured that the following inequality is true \begin{equation} n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!} . \end{equation} Anyhow I could neither prove nor disprove it. I could only check, by using Stirling's Formula, that the ratio of the right and left members tends to 1 as $n \rightarrow \infty$. Any help is welcome.

Answer 1

$$ \begin{align} \log\left(n\left(1+\frac1n\right)^{-n}\right) &=\log(n)-n\log\left(1+\frac1n\right)\\ &=\log(n)+n\log\left(1-\frac1{n+1}\right)\\ &\le\log(n)-\frac{n}{n+1}\\ &\le\log(n)-\frac{n-1}n\\ &=\frac1n\int_1^n\log(x)\,\mathrm{d}x\\ &\le\frac1n\sum_{k=2}^n\log(k)\\ &=\log\left(\sqrt[n]{n!}\right) \end{align} $$ Exponentiate and rearrange to get $$ n^{n+1}\le(n+1)^n\sqrt[n]{n!} $$

Answer 2

From $$ e^{-1} = \left(e^{-\frac{1}{n+1}}\right)^{n+1} > \left(1-\frac{1}{n+1}\right)^{n+1}(\because e^x \ge x+1) $$

and

$$ \left(1-\frac{1}{n+1}\right)^{n+1} \cdot \left(1 + \frac{1}{n}\right)^{n+1} = 1 $$

Note

$$e \le (1+\frac{1}{n})^{n+1}=\frac{(1+n)^{n+1}}{n^{n+1}} \Leftrightarrow n^{n+1} \le \frac{(n+1)^{n+1}}{e} $$

Also note $$n!e^n=\sum_{k=0}^{\infty}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\binom{n}{k}}= ({n+1})^n \Leftrightarrow \sqrt[n]{n!} \ge \frac{n+1}{e} $$

This gives us that $$(n+1)^{n} \sqrt[n]{n!} \ge \frac{(n+1)^{n+1}}{e} \ge n^{n+1}$$.

Answer 3

If you feel inclined to do "nice" (erm) calculus, you can always choose to study one of these two functions: $$ f\colon x\in [1,\infty) \mapsto \Gamma(x+1) (x+1)^{x^2} - x^{x^2+x} $$ or $$ g\colon x\in [1,\infty) \mapsto \frac{\Gamma(x+1) (x+1)^{x^2}}{x^{x^2+x}} $$ and show respectively that $f\geq 0$ or $g \leq 1$ on their domain. (for instance, as $g$ is increasing, if I'm not mistaken, so after showing this you would only have to compute $\lim_\infty g$).

This will not be very enjoyable, but should work: whichever of the two functions you choose, it'll be differentiable and nicely behaved. (If you want, you can even define them on $[2,\infty)$ instead if technicalities at $1$ pop up.)

Answer 4

Taking logarithms the inequality is equivalent to $$ (n+1)\log n\le n\log(n+1)+\frac1n\sum_{k=1}^n\log k. $$ We estimate now the sum: $$ \sum_{k=1}^n\log k\ge\int_1^n\log x\,dx=n\log n-n+1. $$ It will be enough to prove that $$ n\log n\le n\log(n+1)-1+\frac1n, $$ or equivalently $$ 1-\frac1n\le n\log\Bigl(1+\frac1n\Bigr). $$ This follows from the inequality $\log(1+x)\ge x-x^2/2$, $0<x<1$.