Considering the constant coefficient $n-1$ appearing in the problem, we apply the Cauchy-Schwarz inequality as follows:
$$\begin{aligned} \left(\sum_{i=1}^{n} a_{i}\right)^{2} & =\left[\left(a_{1}+a_{2}\right)+a_{3}+a_{4}+\cdots+a_{n}\right]^{2} \\ & \leqslant \underbrace{(1+1+\cdots+1)}_{n-1}\left[\left(a_{1}+a_{2}\right)^{2}+a_{3}^{2}+a_{4}^{2}+\cdots+a_{n}^{2}\right] \\ & =(n-1)\left(\sum_{i=1}^{n} a_{i}^{2}+2 a_{1} a_{2}\right) . \end{aligned}$$

From the problem statement, we have $A < -\left(\sum_{i=1}^{n} a_{i}^{2}\right) + \frac{1}{n-1}\left(\sum_{i=1}^{n} a_{i}\right)^{2}$
$$\leqslant -\left(\sum_{i=1}^{n} a_{i}^{2}\right) + \left(\sum_{i=1}^{n} a_{i}^{2} + 2 a_{1} a_{2}\right) = 2 a_{1} a_{2},$$

Thus, $A < 2 a_{1} a_{2}$.
Similarly, for $1 \leqslant i < j \leqslant n$, we have $A < 2 a_{i} a_{j}$.