I check two case If $ a_{1}a_{2} + a_{2}a_{3} + \cdots + a_{n}a_{1}\ge \frac {1}{n}$ we have $ \frac {a_{1}}{a_{2}^2 + a_{2}} + \frac {a_{2}}{a_{3}^2 + a_{3}} + \cdots + \frac {a_{n}}{a_{1}^2 + a_{1}}\ge \frac {n^2}{n + 1}$ since AMGM inequality and If $ a_{1}a_{2} + a_{2}a_{3} + \cdots + a_{n}a_{1}\le \frac {1}{n}$ Use Cauchy Schwarz inequality $ RHS \ge \frac {a_1}{\sqrt {a_2 + 1}} + \cdots + \frac {a_n}{\sqrt {a_1 + 1}}\ge \frac {(a_1 + \cdots + a_n)^2}{a_1\sqrt {a_2 + 1} + \cdots + a_n\sqrt {a_1 + 1}}$ And now $ (a_1\sqrt {a_2 + 1} + \cdots + a_n\sqrt {a_1 + 1})^2\le (a_1 + \cdots + a_n)(a_{1}a_{2} + a_{2}a_{3} + \cdots + a_{n}a_{1} + a_1 + \cdots + a_n) \le \frac {n + 1}{n}$ we have done

This is another method First denote $ T= \sum_{i=1}^{n} \frac {a_{i}}{a_{i+1}}, (a_{n+1}=a_{1})$, then applying AM-GM Inequality we get $ T \ge n$. Next applying Caughy Inequality we get $ \sum_{i=1}^{n} \frac {a_{i}}{a_{i+1}^2+a_{i+1}}\ge \frac {T^2}{\sum_{i=1}^n \frac {a_{i}(a_{i+1}+1)}{a_{i+1}}}= \frac {T^2}{T+1} \ge \frac {nT}{n+1} (T \ge n)$ Finally applying Caughy again we have $ (\sum_{i=1}^n a_{i}a_{i+1})(\sum_{i=1}^{n} \frac {a_{i}}{a_{i+1}^2+a_{i+1}}) \ge (\sum_{i=1}^n a_{i}a_{i+1})(\frac {nT}{n+1}) \ge \frac {n}{n+1}(\sum_{i=1}^n a_{i})^2)=\frac {n}{n+1}$ Q.E.D.

Allnames wrote: If $ a_{1}a_{2} + a_{2}a_{3} + \cdots + a_{n}a_{1}\le \frac {1}{n}$ Another method: Using $\text{Holder inequality,}$ $$\begin{aligned} \left(a_{1}a_{2} + a_{2}a_{3} + \cdots + a_{n}a_{1}\right)\left(\frac {a_{1}}{a_{2}^2 + a_{2}} + \frac {a_{2}}{a_{3}^2 + a_{3}} + \cdots + \frac {a_{n}}{a_{1}^2 + a_{1}}\right) &\geqslant\dfrac{(a_1+a_2+\cdots +a_n)^3}{a_1(a_2+1)+a_2(a_3+1)+\cdots +a_n(a_1+1)}\\ &=\frac{1}{1+(a_1a_2+a_2a_3+\cdots +a_na_1)}\geqslant\frac {n}{n + 1}.\blacksquare\end{aligned}$$