(2) By symmetry, without loss of generality, assume $a_{n}0
\end{array}$$$$a_{1}+a_{2}+\cdots+a_{n-1}>a_{1}+a_{2}+\cdots+a_{n} \geqslant \sqrt{(n-1)\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)}$$Squaring both sides, we get$$\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)^{2}>(n-1)\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) Therefore, applying the conclusion of (1) to $n-1$, we have (n-1)\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n-1}^{2}\right) \geqslant\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)^{2}$$$$

From (1) and (2), we get $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n-1}^{2}>a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$, which implies $a_{n}^{2}<0$. This is a contradiction, hence $a_{n} \geqslant 0$. Similarly, $a_{1}, a_{2}, \cdots, a_{n-1} \geqslant 0$.