Find all real sequences x1, x2, ... , x1995 which satisfy 2√(xn - n + 1) ≥ xn+1 - n + 1 for n = 1, 2, ... , 1994, and 2√(x1995 - 1994) ≥ x1 + 1.
Solution
Answer: the only such sequence is 1, 2, 3, ... , 1995.
Put x1995 = 1995 + k. We show by induction (moving downwards from 1995) that xn ≥ n + k. For suppose xn+1 ≥ n + k + 1, then 4(xn - n + 1) ≥ (xn+1- n + 1)2 ≥ (k+2)2 ≥ 4k + 4, so xn ≥ n + k. So the result is true for all n ≥ 1. In particular, x1 ≥ 1 + k. Hence 4(x1995 - 1994) = 4(1 + k) ≥ (2 + k)2 = 4 + 4k + k2, so k2 ≤ 0, so k = 0.
Hence also xn ≥ n for n = 1, 2, ... , 1994. But now if xn = n + k, with k > 0, for some n < 1995, then the same argument shows that x1 ≥ 1 + k and hence 4 = 4(x1995 - 1994) ≥ (x1 + 1)2 ≥ (2 + k)2 = 4 + 4k + k2 > 4. Contradiction. Hence xn = n for all n ≤ 1995.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002