已知正项数列 $\{a_n\}$，满足 $a_1=1$，$a_{n+1}=\dfrac{a_n}{\sqrt{a_n^2+1}}$，求证：$$\ln(n+1)<a_1a_2+a_2a_3+\cdots+a_na_{n+1}<\ln\left(\dfrac{2n}3+1\right)+\dfrac12.$$
解析
根据题意，有\[\dfrac{1}{a_{n+1}^2}=1+\dfrac{1}{a_n^2},\]于是\[\dfrac{1}{a_n^2}=n,n\in\mathbb N^{\ast},\]进而\[a_n=\dfrac{1}{\sqrt n},n\in\mathbb N^{\ast},\]欲证不等式即\[\ln (n+1)<\sum_{k=1}^n\dfrac{1}{\sqrt{k(k+1)}}<\ln\left(\dfrac {2n}3+1\right)+\dfrac 12.\]分析通项，只需要证明对任意 $n\geqslant 2$，$n\in\mathbb N^{\ast}$，均有\[\ln(n+1)-\ln n<\dfrac{1}{\sqrt{n(n+1)}}<\ln\left(\dfrac{2n}3+1\right)-\ln\left(\dfrac{2(n-1)}3+1\right),\]也即\[\ln\left(1+\dfrac 1n\right)<\dfrac{1}{\sqrt{n(n+1)}}<\ln\left(1+\dfrac{2}{2n+1}\right).\]事实上，根据对数平均不等式，有\[\dfrac{\dfrac 1n-\dfrac{1}{n+1}}{\ln\dfrac{1}{n}-\ln \dfrac{1}{n+1}}>\sqrt{\dfrac{1}{n(n+1)}},\]于是\[\ln\left(1+\dfrac 1n\right)<\dfrac{1}{\sqrt{n(n+1)}},\]因此左侧不等式成立．对于右侧不等式，显然不成立，需要寻找其他方法． 考虑到\[\sum_{k=1}^n\dfrac{1}{\sqrt{k(k+1)}}<\dfrac 12\sum_{k=1}^n\left(\dfrac 1k+\dfrac1{k+1}\right)=\dfrac 12+\sum_{k=2}^n\dfrac1k+\dfrac1{2(n+1)}<\dfrac 12+\sum_{k=2}^{n=1}\dfrac1k,\]根据对数平均不等式，有\[\dfrac 1n<\ln\dfrac{2n+1}{2n-1},\]因此\[\sum_{k=1}^n\dfrac{1}{\sqrt{k(k+1)}}<\dfrac 12+\ln\dfrac{2n+3}{3},\]右边不等式成立． 综上所述，原不等式得证．