Set $b_n = \frac{1}{a_n^2}$. Then we have $b_1=1$, $b_n(n^2+1) = b_{n-1}(n-1)^2$. We wish to prove that $\sum b_1 \leq 1 + \sqrt{1-b_n}$. Note that $b_n = b_{n-1}(n-1)^2 - b_n(n^2)$ so the sum $b_1 + b_2 + \cdots + b_n$ telescopes: \begin{eqnarray*} \sum_{i=1}^{n} b_i &=& b_1 + b_1 \cdot 1^2 - b_2 \cdot 2^2 + b_2 \cdot 2^2 - b_3 \cdot 3^2 \cdots + b_{n-1}(n-1)^2 - b_n(n^2) \\ &=& 2 - b_n \cdot n^2\end{eqnarray*} It then suffices to show that $1-b_n \cdot n^2 \leq \sqrt{1-b_n}$. Note that $0 < b_n \leq 1 \Longrightarrow 0 \leq 1-b_n < 1$. We have $1-b_n \cdot n^2 \leq 1 - b_n \leq \sqrt{1-b_n}$ so yay.