IMO 2004 SL A1

Origin: KOR | Category: Algebra

Problem

Let n \geq3 be an integer and t1, t2, . . . , tn positive real numbers such that n2 + 1 > (t1 + t2 + \cdot \cdot \cdot + tn)  1 t1

• 1 t2
• \cdot \cdot \cdot + 1 tn  . Show that ti, tj, tk are the side lengths of a triangle for all i, j, k with 1 \leqi < j < k \leqn.

Solution

By symmetry, it is enough to prove that t1 + t2 > t3. We have

n  i=1 ti

n  i=1 t i

= n2 +  i<j  ti tj

• tj ti −2  . (1) All the summands on the RHS are positive, and therefore the RHS is not smaller than n2 + T , where T = (t1/t3 + t3/t1 −2) + (t2/t3 + t3/t2 −2). We note that T is increasing as a function in t3 for t3 \geqmax{t1, t2}. If t1+t2 = t3, then T = (t1+t2)(1/t1+1/t2)−1 \geq3 by the Cauchy–Schwarz inequality. Hence, if t1 + t2 \leqt3, we have T \geq1, and consequently the RHS in (1) is greater than or equal to n2 + 1, a contradiction. Remark. In can be proved, for example using Lagrange multipliers, that if n2 + 1 in the problem is replaced by (n + \sqrt 10 −3)2, then the statement remains true. This estimate is the best possible.