International Mathematical Olympiad

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 $$\begin{array}{rl}& (t_1+\cdots +t_n)\left(\frac{1}{t_1}+\cdots +\frac{1}{t_n}\right)\\ & =\sum _{1\le i<j\le n}\left[\frac{t_i}{t_j}+\frac{t_j}{t_i}\right]+n\\ & =\frac{t_1}{t_3}+\frac{t_3}{t_1}+\frac{t_2}{t_3}+\frac{t_3}{t_2}+\sum _{\begin{array}{c}1\le i<j\le n\\ (i,j)\notin \{(1,3),(2,3)\}\end{array}}[\frac{t_i}{t_j}+\frac{t_j}{t_i}]+n\\ & \ge \frac{t_1+t_2}{t_3}+t_3\left(\frac{1}{t_1}+\frac{1}{t_2}\right)+\sum _{\begin{array}{c}1\le i<j\le n\\ (i,j)\notin \{(1,3),(2,3)\}\end{array}}2+n\\ & \ge \frac{t_1+t_2}{t_3}+\frac{4t_3}{t_1+t_2}+2(C_n^2-2)+n\\ & =4\frac{t_3}{t_1+t_2}+\frac{t_1+t_2}{t_3}+n^2-4.\end{array}(1)$$

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.