### IMO 1982

Problem A3

Consider infinite sequences {xn} of positive reals such that x0 = 1 and x0 ≥= x1 ≥ x2 ≥ ... .

(a)  Prove that for every such sequence there is an n ≥ 1 such that:

x02/x1 + x12/x2 + ... + xn-12/xn ≥ 3.999.

(b)  Find such a sequence for which:

x02/x1 + x12/x2 + ... + xn-12/xn < 4   for all n.

Solution

(a)  It is sufficient to show that the sum of the (infinite) sequence is at least 4. Let k be the greatest lower bound of the limits of all such sequences. Clearly k ≥ 1. Given any ε > 0, we can find a sequence {xn} with sum less than k + ε. But we may write the sum as:

x02/x1 + x1( (x1/x1)2/(x2/x1) + (x2/x1)2/(x3/x1) + ... + (xn/x1)2/(xn+1/x1) + ... ).

The term in brackets is another sum of the same type, so it is at least k. Hence k + ε > 1/x1 + x1k. This holds for all ε > 0, and so k ≥ 1/x1 + x1k. But 1/x1 + x1k ≥ 2√k, so k ≥ 4.

(b)  Let xn = 1/2n. Then x02/x1 + x12/x2 + ... + xn-12/xn = 2 + 1 + 1/2 + ... + 1/2n-2 = 4 - 1/2n-2 < 4.

Solutions are also available in     Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.