Nice problem Denote the segments on the line by $\mathcal{I}_k$ for $1\le k\le n+1$ and let $\{a_k, b_k\}$ be the respective endpoints. Consider the following segments $$\{a_{1}a_{2}, a_{2}a_{3}, \dots, a_{n}a_{n+1}, b_{1}b_{2}, \dots, b_{n}b_{n+1}\}$$and take the one with minimal length, wlog let it be $a_{t}a_{t+1}$ with length $l$ . Now that fraction starts to make sense, consider the segments $a_{t}b_{t}, a_{t+1}b_{t+1}$ and easily observe that the segment at their intersection has length of at least $l(n-k)+l(k-1)=l(n-1)$ which is at least $\frac{n-1}{n}$ of the length of $a_{t}b_{t}$ as desired.