Given an integer $n \geqslant 3$. Find the maximum positive number $\lambda$, such that if positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}<\lambda\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{3} \text {, }
$$

then $a_{1}, a_{2}, a_{3}$ can definitely be the lengths of the sides of some triangle.
Solve $\lambda_{\max }=\frac{10}{[8+\sqrt{10}(n-3)]^{2}}$.
(1) If $\lambda>\frac{10}{[8+\sqrt{10}(n-3)]^{2}}$, take
$$
\begin{array}{l}
a_{1}=4, a_{2}=a_{3}=2, \\
a_{4}=a_{5}=\cdots=a_{n}=\sqrt{10} .
\end{array}
$$

Then the inequality holds, but $a_{1} 、 a_{2} 、 a_{3}$ cannot form a triangle.
(2) If $\lambda=\frac{8}{[8+\sqrt{10}(n-3)]^{2}}$, and the inequality holds, prove by contradiction that $a_{1}<a_{2}+a_{3}$.
Assume $a_{1} \geqslant a_{2}+a_{3}$. Then
$$
\begin{array}{l}
\left(a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}\right) \cdot \\
{\left[4^{3}+2^{3}+2^{3}+(\sqrt{10})^{3}+(\sqrt{10})^{3}+\cdots+(\sqrt{10})^{3}\right]^{2}} \\
\geqslant\left(16 a_{1}+4 a_{2}+4 a_{3}+10 a_{4}+\cdots+10 a_{n}\right)^{3} \\
\geqslant\left(10 a_{1}+10 a_{2}+10 a_{3}+10 a_{4}+\cdots+10 a_{n}\right)^{3},
\end{array}
$$

i.e., $a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3} \geqslant \lambda\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{3}$, a contradiction.
Thus, $a_{1}<a_{2}+a_{3}$.
Similarly, $a_{2}<a_{1}+a_{3}, a_{3}<a_{1}+a_{2}$.
Hence, $a_{1} 、 a_{2} 、 a_{3}$ can be the lengths of the three sides of a triangle.