Will write more later, have to sleep now

We have to show that $$\left(\sum_{i=1}^n\frac{a_i}{a_{i+1}}\right)-n \leq \frac{1}{2a_1a_n}\sum_{i=1}^n(a_i-a_{i+1})^2.$$We induct on $n$. For $n=2$, we have $$\frac{a_1}{a_2}+\frac{a_2}{a_1}-2=\frac{a_1^2+a_2^2-2a_1a_2}{a_1a_2}.$$Now suppose that $$\left(\sum_{i=1}^{n-1}\frac{a_i}{a_{i+1}}\right)-(n-1) \leq \frac{1}{2a_1a_{n-1}}\sum_{i=1}^{n-1}(a_i-a_{i+1})^2.$$Then \begin{align*} \left(\sum_{i=1}^n\frac{a_i}{a_{i+1}}\right)-n&=\left(\sum_{i=1}^{n-1}\frac{a_i}{a_{i+1}}\right)-(n-1)+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}-\frac{a_{n-1}}{a_1}-1\\&\leq\frac{1}{2a_1a_{n-1}}\sum_{i=1}^{n-1}(a_i-a_{i+1})^2+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}-\frac{a_{n-1}}{a_1}-1 \end{align*}We now have to show $$\frac{1}{2a_1a_{n-1}}\sum_{i=1}^{n-1}(a_i-a_{i+1})^2+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}-\frac{a_{n-1}}{a_1}-1 \leq \frac{1}{2a_1a_n}\sum_{i=1}^n(a_i-a_{i+1})^2.$$Notice that $$\sum_{i=1}^n(a_i-a_{i+1})^2=\sum_{i=1}^n(a_i^2-a_ia_{i+1})$$Thus we have to show that $$a_n\left(\sum_{i=1}^{n-1}a_i^2-\sum_{i=1}^{n-1}a_ia_{i+1}\right)+a_1a_{n-1}^2+a_{n-1}a_n^2-a_{n}a_{n-1}^2-a_1a_{n-1}a_n \leq a_{n-1}\sum_{i=1}^n(a_i^2-a_ia_{i+1})$$which is equivalent to $$a_n\left(\sum_{i=1}^{n-1}(a_i-a_{i+1})^2\right)\leq a_{n-1}\left(\sum_{i=1}^{n-1}(a_i-a_{i+1})^2\right)$$which is true.

to ACGNmath: there seems to be a mistake on your line 14. it should be (sum)(i=1)(n)(ai-ai+1)^2=2*(sum)(i=1)(n)(ai^2-ai*ai+1) also, during induction, it shouldn't be stuff like sum(i=1)(n-1)(ai-ai+1)^2 it should be (sum(i=1)(n-2)(ai-ai+1)^2)+(an-1-a1)^2 as it does not cycle during induction