Let $a_{i+1}-a_{i}=h$, then $\frac{1}{a_{k}}\le \frac{2a_{k}}{2(a_{k}^{2}-h^{2})}=\frac{1}{2}(\frac{1}{a_{k-1}}+\frac{1}{a_{k+1}})$. Therefore $\frac{1}{a_{k}^{2}}\le \frac{1}{2}(\frac{1}{a_{k}a_{k-1}}+\frac{1}{a_{k}a_{k+1}})=\frac{h}{2}(\frac{1}{a_{k-1}}-\frac{1}{a_{k+1}})$. It give \[\sum_{k=2}^{n}\frac{1}{a_{k}^{2}}\le \frac{h}{2}(\frac{1}{a_{1}}+\frac{1}{a_{2}}-\frac{1}{a_{n}}-\frac{1}{a_{k+1}})=\frac{n-1}{2}(\frac{1}{a_{1}a_{n}}+\frac{1}{a_{2}}{a_{n+1}}). \]