9. We arrange the given $n$ segments in length as $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. If no 3 segments can form a triangle, then we have $a_{3} \geqslant a_{1}+a_{2}, a_{6} \geqslant a_{2}+a_{3}, \cdots, a_{n} \geqslant a_{n-2} +a_{n-1}$, thus $a_{n} \geqslant a_{n-2}+a_{n-3}+\cdots+a_{3}+2 a_{2}+a_{1}$, hence $a_{n}>\sum_{j=1}^{n-2} a_{j}$, which means $a_{n}, a_{n-2}, a_{n-3}, \cdots, a_{3}, a_{2}, a_{1}$ these $n-1$ segments (excluding $a_{n-1}$) cannot form an $(n-1)$-gon, leading to a contradiction.