Question 18
For $n \ge 2$, let $a_1, a_2, \dots, a_n, a_{n+1}$ be positive and $a_2 - a_1 = a_3 - a_2 = \dots = a_{n+1} - a_n \ge 0$. Prove that
$$
\frac{1}{a_2^2} + \frac{1}{a_3^2} + \dots + \frac{1}{a_n^2} \le \frac{n-1}{2} \cdot \frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}}
$$
Determine when equality holds.
$$
\frac{1}{a_2^2} + \frac{1}{a_3^2} + \dots + \frac{1}{a_n^2} \le \frac{n-1}{2} \cdot \frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}}
$$
Determine when equality holds.