Proof (1) From the problem, we get
$$\begin{array}{l} \left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right)>0 \text {, i.e., } \\ (a+b+c)(a+b-c)(a-b+c)(b+c-a)>0, \end{array}$$

Since $a, b, c > 0$, at least three of the factors on the left side of the inequality are positive. Since the product of the four factors is positive, all four factors must be positive. Therefore, $a+b>c$, $b+c>a$, and $c+a>b$, which means $a, b, c$ must be the lengths of the sides of some triangle.
(2) Using the Cauchy-Schwarz inequality, we get
$$\begin{aligned} ( & n-1)\left(a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4}\right)<\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)^{2} \\ = & {\left[1 \times \frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+1 \times \frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+1 \times a_{4}^{2}+\cdots+1 \times a_{n}^{2}\right]^{2} } \\ \leqslant & (\underbrace{1^{2}+1^{2}+\cdots+1^{2}}_{(n-1) \uparrow})\left[\frac{1}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}+\frac{1}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}\right. \\ & \left.+a_{4}^{4}+\cdots+a_{n}^{4}\right] \\ = & (n-1)\left[\frac{1}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}+\frac{1}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}+a_{4}^{4}+\cdots+a_{n}^{4}\right], \end{aligned}$$

Thus, $2\left(a_{1}^{4}+a_{2}^{4}+a_{3}^{4}\right)<\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}$.
From (1), we know that $a_{1}, a_{2}, a_{3}$ are the lengths of the sides of some triangle. By symmetry, any three numbers among $a_{1}, a_{2}, \cdots, a_{n}$ can be the lengths of the sides of some triangle.