Prove by (1) and the Cauchy-Schwarz inequality:
$$\begin{array}{l} (n-1)\left(a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4}\right)<\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)^{2}= \\ \left(\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+a_{4}^{2}+\cdots+a_{n}^{2}\right)^{2} \leqslant \\ (n-1)\left[\frac{\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}}{4}+\frac{\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}}{4}+a_{4}^{4}+\cdots+a_{n}^{4}\right] \end{array}$$

Thus, we obtain $2\left(a_{1}^{4}+a_{2}^{4}+a_{3}^{4}\right)<\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)^{2}$.
The ingenuity of this proof method lies in transforming the sum of three terms into the sum of two terms, thereby converting an expression that is the sum of $n$ terms into an expression that is the sum of $n-1$ terms. When applying the Cauchy-Schwarz inequality, a factor of $(n-1)$ appears, which cancels out with the same factor on the other side, thus solving the problem with just one application of the Cauchy-Schwarz inequality.