IMO 2004 · Problem 4

Let \(n \geq 3\) be an integer. Let \(t_1,t_2,\ldots,t_n\) be positive real numbers such that \[ n^2+1>(t_1+t_2+\ldots+t_n) \left(\frac1{t_1}+\frac1{t_2}+\ldots+\frac1{t_n}\right). \] Show that \(t_i,t_j,t_k\) are side lengths of a triangle for all \(i,j,k\) with \(1\leq i<j<k\leq n\).

Official text of the problem. Official problem paper (PDF)

Solution

Expanding the product gives \[ \left(\sum_{i=1}^n t_i\right)\left(\sum_{i=1}^n\frac1{t_i}\right) =n^2+\sum_{1\leq i<j\leq n}\frac{(t_i-t_j)^2}{t_it_j}. \] Thus the hypothesis implies \[ \sum_{1\leq i<j\leq n}\frac{(t_i-t_j)^2}{t_it_j}<1. \tag{1} \]

Suppose, for a contradiction, that some three of the numbers do not form a triangle. Since they are positive, we may label them \(a,b,c\) so that \(c\geq a+b\).

Set \(s=a+b\) and \(r=c/s\geq1\). Since \(ab\leq s^2/4\), we have \[ \frac1a+\frac1b=\frac{s}{ab}\geq\frac4s. \] Consequently, \[ \begin{aligned} (a+b+c)\left(\frac1a+\frac1b+\frac1c\right) &\geq(s+c)\left(\frac4s+\frac1c\right)\\ &=(1+r)\left(4+\frac1r\right)\\ &=10+\frac{(r-1)(4r-1)}r\\ &\geq10. \end{aligned} \] Applying the initial identity to these three numbers therefore yields \[ \frac{(a-b)^2}{ab} +\frac{(a-c)^2}{ac} +\frac{(b-c)^2}{bc} =(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)-9 \geq1. \] These are three terms of the sum in (1), and every other term is nonnegative. This contradicts (1).

Hence every three of the given numbers satisfy the strict triangle inequalities and are side lengths of a triangle. \(\square\)

The solution was checked against Evan Chen's published IMO solution notes.