The claim is that
$$\frac{\sum_{1 \le i < j \le n} a_i a_j}{\binom{n}{2}} \le \left( \frac{\sum_{k=1}^{n} a_k}{n} \right)^2 . \text{ Equivalently,}$$
$$2n \sum_{1 \le i < j \le n} a_i a_j \le (n-1) \left( \sum_{k=1}^{n} a_k^2 + 2 \sum_{1 \le i < j \le n} a_i a_j \right), \text{ or}$$
$$2 \sum_{1 \le i < j \le n} a_i a_j \le (n-1) \sum_{k=1}^{n} a_k^2, \text{ i.e.,}$$
$$\left( \sum_{k=1}^{n} a_k \right)^2 \le n \sum_{k=1}^{n} a_k^2,$$
which is true by the Cauchy-Schwarz inequality. There is equality iff $a_1 = a_2 = \dots = a_n$.