How to prove $\sqrt[n]{n}>\sqrt[n+1]{n+1}$ ,$n\ge 3$ monotone decreases by using this hint? I can solve it in other ways but I don't know how to solve it using this hint.
hints: Consider the expression $(1+n)^{n}\cdot n^{\frac{n+1}{n}}$ and use am-gm inequality.
note that$3^{\frac{-1}{3}}+\frac{1}{4}<1$
Your inequality $$\sqrt[n]{n}>\sqrt[n+1]{n+1}$$ for $n\geq 3$ is equivalent to $$\left(1+\frac{1}{n}\right)^n<n$$ to show this we need, that $$\binom{n}{k}\le \frac{n^k}{n!}$$ and $$\frac{1}{k!}\le \frac{1}{2^{k-1}}$$ with the help of these inequalities we get $$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\cdot \frac{1}{n^k}=2+\sum_{k=2}^n\binom{n}{k}\cdot \frac{1}{n^k}\le 2+\sum_{k=2}^n\frac{n^k}{k!}\cdot \frac{1}{n^k}=2+\sum_{k=2}^n\frac{1}{k!}\le 2+\sum_{k=2}^n\frac{1}{2^{k-1}}<3$$
