Assume on the contrary that there exist three numbers among $t_1, t_2, \dots, t_n$ that do not form the sides of a triangle. Without loss of generality, we may assume that these three numbers are $t_1, t_2, t_3$, and $t_1+t_2 \le t_3$. One has
$$\begin{aligned} & (t_1+\cdots+t_n)\left(\frac{1}{t_1}+\cdots+\frac{1}{t_n}\right) \\ &=\sum_{1\le i<j\le n}\left[\frac{t_i}{t_j}+\frac{t_j}{t_i}\right]+n \\ &=\frac{t_1}{t_3}+\frac{t_3}{t_1}+\frac{t_2}{t_3}+\frac{t_3}{t_2}+\sum_{\substack{1\le i<j\le n \\ (i,j)\notin\{(1,3), (2,3)\}}}[\frac{t_i}{t_j}+\frac{t_j}{t_i}]+n \\ &\ge\frac{t_1+t_2}{t_3}+t_3\left(\frac{1}{t_1}+\frac{1}{t_2}\right)+\sum_{\substack{1\le i<j\le n \\ (i,j)\notin\{(1,3), (2,3)\}}}2+n \\ &\ge\frac{t_1+t_2}{t_3}+\frac{4t_3}{t_1+t_2}+2(C_n^2-2)+n \\ &=4\frac{t_3}{t_1+t_2}+\frac{t_1+t_2}{t_3}+n^2-4. \end{aligned} \quad (1)$$

If $x = \frac{t_3}{t_1+t_2}$, then $x \ge 1$, and $4x + \frac{1}{x} - 5 = \frac{(x-1)(4x-1)}{x} \ge 0$.
Together with (1), we obtain that
$$(t_1+\cdots+t_n)\left(\frac{1}{t_1}+\cdots+\frac{1}{t_n}\right) \ge 5+n^2-4 = n^2+1,$$
a contradiction. This completes the proof.