\begin{align*} \sum_{i=1}^{2016}\frac{1}{a_{i+1}^2+a_{i+2}^2} &\leq\sum_{i=1}^{2016}\frac{1}{(i+1007)^2+(i+1008)^2}\\ &=\sum_{i=1}^{2016}\frac{1}{2i^2+4030i+2030113}\\ &<\sum_{i=1}^{2016}\frac{1}{2i^2+4030i+2030112}\\ &=\sum_{i=1}^{2016}\frac{1}{2(i+1007)(i+1008)}\\ &=\sum_{i=1}^{2016}\left(\frac{1}{2(i+1007)}-\frac{1}{2(i+1008)}\right)\\ &<\frac{1}{2016} \end{align*}

bgn wrote: The sequence $(a_n)$ is defined as: $$a_1=1007$$$$a_{i+1}\geq a_i+1$$Prove the inequality: $$\frac{1}{2016}>\sum_{i=1}^{2016}\frac{1}{a_{i+1}^{2}+a_{i+2}^2}$$ This conclusion is also true The sequence $\{ {a_n}\} $is defined as: ${a_1} = m,{a_{i + 1}} \ge {a_i} + 1(i = 1,2,3, \cdots ),m \in N,k \in N,k \ge 1$. Prove the inequality: $$\sum\limits_{i = 1}^k {\frac{1}{{a_{i + 1}^2 + a_{i + 2}^2}}} < \frac{1}{{2\left( {m + 1} \right)}}$$Proof $$\sum\limits_{i = 1}^k {\frac{1}{{a_{i + 1}^2 + a_{i + 2}^2}}} \le \sum\limits_{i = 1}^k {\frac{1}{{2{a_{i + 1}}{a_{i + 2}}}}} \le \frac{1}{2}\sum\limits_{i = 1}^k {\frac{{{a_{i + 2}} - {a_{i + 1}}}}{{{a_{i + 1}}{a_{i + 2}}}}} = \frac{1}{2}\sum\limits_{i = 1}^k {\left( {\frac{1}{{{a_{i + 1}}}} - \frac{1}{{{a_{i + 2}}}}} \right)} $$$$ = \frac{1}{2}\left( {\frac{1}{{{a_2}}} - \frac{1}{{{a_{k + 2}}}}} \right) < \frac{1}{{2{a_2}}} \le \frac{1}{{2(m + 1)}}$$

we have $ \frac{1}{a_{i+1}^2 + a_{i+2}} < \frac{1}{2a_{i+1}a_{i+2}} \le \frac{1}{2} (\frac{1}{a_{i+1}} - \frac{1}{a_{i+2}}) $ So $ RHS < \frac{1}{2}(\frac{1}{a_2} - \frac{1}{a_{2017}}) <\frac{1}{2} .\frac{1}{a_2}\le \frac{1}{2016} $ .

It's easy to see that $a_i\geqslant 1006+i$ for all $i\geqslant 1.$ So, \begin{align*} \sum_{i=1}^{2016}\frac{1}{a_{i+1}^2+a_{i+2}^2}&\leqslant \sum_{i=1}^{2016}\frac{1}{(1007+i)^2+(1008+i)^2}\\ &\leqslant \frac{1}{2}\sum_{i=1}^{2016} \frac{1}{(1007+i)(1008+i)}\\ &=\frac{1}{2}\sum_{i=1}^{2016} \left(\frac{1}{1007+i}-\frac{1}{1008+i}\right) \\ &=\frac{1}{2}\left(\frac{1}{1008}-\frac{1}{1008+2016}\right) \\ &< \frac{1}{2016}. \quad \blacksquare \end{align*}

$\sum\limits_{i = 1}^{2016} {\frac{1}{{a_{i + 1}^2 + a_{i + 2}^2}}} \le \sum\limits_{i = 1}^{2016} {\frac{1}{{2{a_{i + 1}}{a_{i + 2}}}}} \le \frac{1}{2}\sum\limits_{i = 1}^{2016} {\frac{{{a_{i + 2}} - {a_{i + 1}}}}{{{a_{i + 1}}{a_{i + 2}}}}} = \frac{1}{2}\sum\limits_{i = 1}^{2016} {\left( {\frac{1}{{{a_{i + 1}}}} - \frac{1}{{{a_{i + 2}}}}} \right)} = \frac{1}{2}(\frac{1}{a_{2}} - \frac{1}{a_{2018}}) \le \frac{1}{2}(\frac{1}{1008} - \frac{1}{a_{2018}}) \le \frac{1}{2016} $