The sequence $(a_n)$ is defined by $a_1=1,a_2=\frac{1}{2}$,$$n(n+1) a_{n+1}a_{n}+na_{n}a_{n-1}=(n+1)^2a_{n+1}a_{n-1}(n\ge 2).$$Prove that $$\frac{2}{n+1}<\sqrt[n]{a_n}<\frac{1}{\sqrt{n}}(n\ge 3).$$
Source: China Nanchang
Tags: inequalities
The sequence $(a_n)$ is defined by $a_1=1,a_2=\frac{1}{2}$,$$n(n+1) a_{n+1}a_{n}+na_{n}a_{n-1}=(n+1)^2a_{n+1}a_{n-1}(n\ge 2).$$Prove that $$\frac{2}{n+1}<\sqrt[n]{a_n}<\frac{1}{\sqrt{n}}(n\ge 3).$$