We consider two cases:
1) Let's first consider the case when $a_1,...,a_n$ fulfill $\sum a_ia_{i+1}\leq \frac{1}{n}.$
By CS we have
\begin{equation}
\left(\sum a_ia_{i+1} \right)\left(\sum\frac{a_i}{a_{i+1}^2+a_{i+1}} \right)\geq \left(\sum a_i \sqrt{\frac{a_{i+1}}{a_{i+1}^2+a_{i+1}}} \right)^2=\left(\sum a_i \frac{1}{\sqrt{a_{i+1}+1}} \right)^2.
\end{equation}
Now observe that the function $f(x)=\frac{1}{\sqrt{x+1}}$ is clearly convex. So by Jensen
\begin{equation}\sum_{i=1}^m\lambda_i f(x_i)\geq f(\sum_{i=1}^m \lambda_i x_i)
\end{equation}
holds for all $x_1,...,x_m>-1$ and all positive $\lambda_1,...,\lambda_m$ with $\sum_{i=1}^m\lambda_i=1.$ Applying this to $\lambda_i=a_i$ and $x_i=a_{i+1}$ we get
\begin{equation}
\left(\sum a_i \frac{1}{\sqrt{a_{i+1}+1}}\right)^2\geq \left(\frac{1}{\sqrt{1+\sum a_ia_{i+1}}}\right)^2=\frac{1}{1+\sum a_ia_{i+1}}.
\end{equation}
So to finish the proof of the inequality all we have to show is
\begin{equation}
n+1\geq n\left(1+\sum a_ia_{i+1} \right)=n+n\sum a_ia_{i+1},
\end{equation}which is true because $\sum a_ia_{i+1}\leq \frac{1}{n}.$
2) Now consider the case where $\sum a_i a_{i+1}\geq \frac{1}{n}.$ Because $a_1,...,a_n$ and $\frac{1}{a_1^2+a_1},...,\frac{1}{a_n^2+a_n}$ are reversly ordered, we obtain by the main theorem on ordered sequences (I hope it's called like that in english) and AM-HM
\begin{equation}
\sum \frac{a_i}{a_{i+1}^2+a_{i+1}}\geq \sum \frac{a_i}{a_i^2+a_i}=\sum \frac{1}{a_i+1}\geq \frac{n^2}{\sum (a_i +1)}=\frac{n^2}{n+1}.
\end{equation}
So in this second case
\begin{equation}
\left( \sum a_ia_{i+1}\right)\left( \sum \frac{a_i}{a_{i+1}^2+a_{i+1}}\right)\geq \frac{1}{n}\frac{n^2}{n+1}=\frac{n}{n+1}.
\end{equation}
Observe that we don't have to prove $\sum a_ia_{i+1}\geq \frac{1}{n}$ or $\leq \frac{1}{n}$ for any such $n$ numbers $a_1,...,a_n$ with $\sum a_i=1$, but that depending on how large $\sum a_ia_{i+1}$ is, we find a proof for the inequality.
