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

Question

Let $a_{1},a_{2},\cdots,a_{2n-1},a_{2n}$be real numbers,I conjecture

$$a_{1}a_{2}a_{3}\cdots a_{2n}=\prod_{i=1}^{2n}a_{i}=\le\left(\dfrac{a_{1}+a_{2}+\cdots+a_{2n}}{2n}\right)^{2n}\tag{1}$$ I have know if $a_{i}$ be postive real numbers,It's AM-GM inequality,but it seem for any real numbers,$(1)$ also hold?right?

basis $n=2$ it is for any real $a_{1},a_{2}$,then have $$a_{1}a_{2}\le\left(\dfrac{a_{1}+a_{2}}{2}\right)^2\Longleftrightarrow 4a_{1}a_{2}\le (a_{1}+a_{2})^2\Longleftrightarrow (a_{1}-a_{2})^2\ge 0$$

Answer 1

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

Answer 2

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 $P = a_1a_2 \cdots a_k$.