May conjecture AM-GM without positivity $a_{1}a_{2}a_{3}\cdots a_{2n} \le\left(\frac{a_{1}+a_{2}+\cdots+a_{2n}}{2n}\right)^{2n}$

This is incorrect. Set $a_1 = a_2 = -1, a_3 = a_4 = 1$.

The case of $2$ numbers does not generalize, because for $k \geq 3$ any pair $(S,P)$ can be attained as the values of $S = a_1 + a_2 + \dots + a_k$ and $P = a_1a_2 \cdots a_k$.