If $a_n=n^\frac 1n-1, n \in \mathbb N$ prove that $0 \le a_n \le \sqrt {2/n}$?

Question

If $a_n=n^{\frac{1}{n}}-1$, $n\in\mathbb{N}$, prove that $0\le a_n\le\sqrt{\frac{2}{n}}$.

I tried with induction and signs, got nowhere. Any help is appreciated.

Answer 1

The inequality holds for $n\in\{1,2\}$. Assume $n\geq 3$. Apply the AM-GM inequality to the sequence of $n$ numbers $$1, 1, \dots, 1, 2, \sqrt{\frac{n}{2}}, \sqrt{\frac{n}{2}}$$ to obtain $$n^{1/n} \leq \frac{n-1+\sqrt{2n}}{n}$$ and your inequality follows.