41st International Mathematical OlympiadIMO 2000 Problems and SolutionsTaejon, KoreaJuly 13-25, 2000 Problem 2A, B, C are positive reals with product 1. Prove that (A - 1 + 1/B) (B - 1 + 1/C) (C - 1 + 1/A) <= 1.SolutionAn elegant solution due to Robin Chapman is as follows: (B - 1 + 1/C) = B(1 - 1/B + 1/(BC)) = B(1 + A - 1/B)). Hence, (A - 1 + 1/B) (B - 1 + 1/C) = B(A^{2} - (1 - 1/B)^{2}) <= B A^{2}. So the square of the product of all three <= B A^{2} C B^{2} A C^{2} = 1. Actually, that is not quite true. The last sentence would not follow if we had some negative left-hand sides, because then we could not multiply the inequalities. But it is easy to deal separately with the case where (A - 1 + 1/B), (B - 1 + 1/C), (C - 1 + 1/A) are not all positive. If one of the three terms is negative, then the other two must be positive. For example, if A - 1 + 1/B < 0, then A < 1, so C - 1 + 1/A > 0, and B > 1, so B - 1 + 1/C > 0. But if one term is negative and two are positive, then their product is negative and hence less than 1. Few people would manage this under exam conditions, but there are plenty of longer and easier-to-find solutions! Back to Competition Problems page |