IMO 1985 SL 18

Problem

6b.(CAN 5) Let x1, x2, . . . , xn be positive numbers. Prove that x2 x2 1 + x2x3 + x2 x2 2 + x3x4

• \cdot \cdot \cdot + x2 n−1 x2 n−1 + xnx1

x2 n x2n + x1x2 \leqn −1. Supplementary Problems

Solution

Set yi = x2 i xi+1xi+2 , where xn+i = xi. Then $n i=1 yi = 1 and the inequality to be proved becomes n i=1 yi 1+yi \leqn −1, or equivalently

n  i=1 1 + yi \geq1. We prove this inequality by induction on n. Since 1+y + 1+y−1 = 1, the inequality is true for n = 2. Assume that it is true for n −1, and let there be given y1, . . . , yn > 0 with $n i=1 yi = 1. Then 1+yn−1 + 1+yn > 1+yn−1yn , which is equivalent to 1 + ynyn−1(1 + yn + yn−1) > 0. Hence by the inductive hypothesis n  i=1 1 + yi \geq n−2  i=1 1 + yi + 1 + yn−1yn \geq1. Remark. The constant n −1 is best possible (take for example xi = ai with a arbitrarily large).