IMO 1978

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Problem B2

{ak} is a sequence of distinct positive integers. Prove that for all positive integers n, ∑1n ak/k2 ≥ ∑1n 1/k.

 

Solution

We use the general rearrangement result: given b1 ≥ b2 ≥ ... ≥ bn, and c1 ≤ c2 ≤ ... ≤ cn, if {ai} is a permutation of {ci}, then ∑ aibi ≥ ∑ cibi. To prove it, suppose that i < j, but ai > aj. Then interchanging ai and aj does not increase the sum, because (ai - aj)(bi - bj) ≥ 0, and hence aibi + ajbj ≥ ajbi + aibj. By a series of such interchanges we transform {ai} into {ci} (for example, first swap c1 into first place, then c2 into second place and so on).

Hence we do not increase the sum by permuting {ai} so that it is in increasing order. But now we have ai > i, so we do not increase the sum by replacing ai by i and that gives the sum from 1 to n of 1/k.

 


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.

 

20th IMO 1978

© John Scholes
jscholes@kalva.demon.co.uk
12 Oct 1998