Given $a_1,a_2,...,a_n>0$, ($n\geq3, n \in \mathbb{N}$), prove that $$\frac{a_{1}^2+a_{2}a_{3}}{a_{1}(a_{2}+a_{3})}+\frac{a_{2}^2+a_{3}a_{4}}{a_{2}(a_{3}+a_{4})}+...+\frac{a_{n-1}^2+a_{n}a_{1}}{a_{n-1}(a_{n}+a_{1})}+\frac{a_{n}^2+a_{1}a_{2}}{a_{n}(a_{1}+a_{2})} \geq n.$$ I have no idea how to approach a problem like this. I have tried and find a somewhat similar problem (but they are written in Chinese and I don't understand a word). Here is the screenshot of the problem and its solution. Similar problem in Chinese
I tried minimizing each term separately $\frac{a_1^2+a_2a_3}{a_1(a_2+a_3)}≥\frac{2a_1\sqrt{a2a3}}{a_1(a_2+a_3)}=\frac{2\sqrt{a_2a_3}}{a_2+a_3}≤1$ (doesn't work) so I thought doing some algebra on all the terms as a whole would help. However, I didn't come up with any methods to do so.
By AM-GM twice we obtain: $$\sum_{i=1}^{n} \frac{a_{i}^2+a_{i+1}a_{i+2}}{a_{i}(a_{i+1}+a_{i+2})}=\sum_{i=1}^{n}\left(\frac{a_{i}^2+a_{i+1}a_{i+2}}{a_{i}(a_{i+1}+a_{i+2})}+1\right)-n=$$ $$=\sum_{i=1}^{n} \frac{(a_i+a_{i+1})(a_i+a_{i+2})}{a_{i}(a_{i+1}+a_{i+2})}-n\geq n\sqrt[n]{\prod_{i=1}^n\frac{(a_i+a_{i+1})(a_i+a_{i+2})}{a_{i}(a_{i+1}+a_{i+2})}}-n\geq$$ $$\geq n\sqrt[n]{\prod_{i=1}^n\frac{(a_i+a_{i+1})\cdot2\sqrt{a_ia_{i+2}}}{a_{i}(a_{i+1}+a_{i+2})}}-n=2n-n=n$$