Taiwan IMO Selection Camp

Using the fact that when $|x| < 1$, $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$, we can deduce that
$$\begin{aligned} \sum_{k=1}^{n} \frac{1}{1-a_k^2} &= \sum_{k=0}^{\infty} \left( \sum_{i=1}^{n} a_i^{2k} \right) = \sum_{k=0}^{\infty} \left( \sum_{i=1}^{n} (a_i^k)^2 \right) \\ &\ge \frac{2}{n-1} \sum_{k=0}^{\infty} \left( \sum_{i<j} a_i^k a_j^k \right) = \frac{2}{n-1} \sum_{k=0}^{\infty} \left( \sum_{i<j} (a_i a_j)^k \right) \\ &= \frac{2}{n-1} \sum_{i<j} \frac{1}{1-a_i a_j}, \end{aligned}$$
where equality holds when $a_i = a_j, 1 \le i < j \le n$. On the other hand, by the Cauchy-Schwarz inequality, we obtain
$$\left( \sum_{i=1}^{n} \frac{1}{1-a_i^2} \right) \left( \sum_{i=1}^{n} (1-a_i^2) \right) \ge n^2.$$
Then from $1 \le \sum_{i=1}^{n} a_i^2 < n$, we obtain
$$\sum_{i=1}^{n} \frac{1}{1-a_i^2} \ge \frac{n^2}{n-1}.$$
where equality holds when $a_1 = a_2 = \dots = a_n = \pm \frac{1}{\sqrt{n}}$. Combining the two inequalities above, we can deduce that
$$\sum_{k=1}^{n} \frac{2}{1-a_k^2} \ge \frac{n^2}{n-1} + \frac{2}{n-1} \sum_{i<j} \frac{1}{1-a_i a_j}.$$

The above is equivalent to
$$(n-1) \sum_{k=1}^{n} \frac{2}{1-a_k^2} \geq \frac{n^2}{2} + \sum_{i<j} \frac{1}{1-a_i a_j},$$
that is,
$$\sum_{i<j} \left( \frac{1}{1-a_i^2} + \frac{1}{1-a_j^2} \right) \geq \frac{n^2}{2} + \sum_{i<j} \frac{1}{1-a_i a_j},$$
that is,
$$\sum_{i<j} \left( \frac{1}{1-a_i^2} + \frac{1}{1-a_j^2} - \frac{1}{1-a_i a_j} \right) \geq \frac{n^2}{2},$$
with equality holding when $a_1 = a_2 = \dots = a_n = \pm \frac{1}{\sqrt{n}}$.