Let $d = a_j - a_{j-1} \ge 0$. If $d > 0$, then
$$\frac{1}{a_k^2} < \frac{1}{a_k^2 - d^2} = \frac{1}{a_{k-1}a_{k+1}} = \frac{a_{k+1} - a_{k-1}}{2d a_{k-1} a_{k+1}} = \frac{1}{2d} \left( \frac{1}{a_{k-1}} - \frac{1}{a_{k+1}} \right)$$
for any $k > 1$. Therefore, we have
$$\begin{aligned} \sum_{k=2}^n \frac{1}{a_k^2} &< \frac{1}{2d} \sum_{k=2}^n \left( \frac{1}{a_{k-1}} - \frac{1}{a_{k+1}} \right) \\ &= \frac{1}{2d} \left( \frac{1}{a_1} - \frac{1}{a_n} + \frac{1}{a_2} - \frac{1}{a_{n+1}} \right) \\ &= \frac{1}{2d} \left( \frac{(n-1)d}{a_1 a_n} + \frac{(n-1)d}{a_2 a_{n+1}} \right) \\ &= \frac{n-1}{2} \cdot \frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}}. \end{aligned}$$
When $d = 0$, let $a_j = c$ for all $j$. Then
$$\sum_{k=2}^n \frac{1}{a_k^2} = \frac{n-1}{c^2}$$
and
$$\frac{n-1}{2} \cdot \frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}} = \frac{n-1}{2} \cdot \frac{2c^2}{c^4} = \frac{n-1}{c^2} = \sum_{k=2}^n \frac{1}{a_k^2}.$$
Therefore, the inequality is proven, and equality holds when $a_1 = a_2 = \dots = a_{n+1}$.