7th APMO 1995 Problem 1 Find all real sequences x 1 , x 2 , ... , x 1995 which satisfy 2√(x n - n + 1) ≥ x n+1 - n + 1 for n = 1, 2, ... , 1994, and 2√(x 1995 - 1994) ≥ x 1 + 1. Solution
: the only such sequence is 1, 2, 3, ... , 1995. Put x 1995 = 1995 + k. We show by induction (moving downwards from 1995) that x n ≥ n + k. For suppose x n+1 ≥ n + k + 1, then 4(x n - n + 1) ≥ (x n+1 - n + 1) 2 ≥ (k+2) 2 ≥ 4k + 4, so x n ≥ n + k. So the result is true for all n ≥ 1. In particular, x 1 ≥ 1 + k. Hence 4(x 1995 - 1994) = 4(1 + k) ≥ (2 + k) 2 = 4 + 4k + k 2 , so k 2 ≤ 0, so k = 0. Hence also x n ≥ n for n = 1, 2, ... , 1994. But now if x n = n + k, with k > 0, for some n  4. Contradiction. Hence x n = n for all n ≤ 1995. 7th APMO 1995 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002