17th Chinese 2002 Problem B3 k is a fixed real number such that 1/2 < k < 1. What is the smallest real h such that if positive reals x 1 ≤ x 2 ≤ ... ≤ x n satisfy (x 1 + 2x 2 + 3x 3 + ... + nx n )/n = k (x 1 + x 2 + ... + x n ), then they also satisfy (x 1 + x 2 + ... + x n ) ≤ h (x 1 + x 2 + ... + x m ), where m = [kn]. Solution Solution by Milivoje Lukic
: smallest h is n/(n+m+1-2kn) for n > 1. For n = 1, there is no positive x 1 such that x 1 /n = k x 1 (with k = (x a + x a+1 + ... + x b ) (a+b)/2. But that is immediate, since x i is an increasing sequence, so a x a + b x b >= (x a + x b ) (a+b)/2, (a+1) x a+1 + (b-1) x b-1 >= (x a+1 + x b-1 ) (a+b)/2 etc. Put S m = ∑ 1 m x i . Then (x 1 + 2 x 2 + ... + m x m ) >= S m (m+1)/2 and ( (m+1) x m+1 + ... + n x n ) >= (S n - S m ) (n+m+1)/2. Adding, (x 1 + 2 x 2 + ... + n x n ) >= (n+m+1)S n /2 - n S m /2. So if (x 1 + 2 x 2 + ... + n x n ) = kn S n , then (n+m+1-2kn) S n  kn, and (since k  kn, so (n+m+1-2kn) > 0. Hence we may divide through to get S n <= h 0 S m , where h 0 = n/(n+m+1-2kn). If we take x 1 = ... = x m = t and x m+1 = ... = x n = t', then with suitable t, t' we can also satisfy kn S n = (x 1 + 2x 2 + ... + nx n ). Hence we will have equality throughout the previous paragraph. Hence h 0 is the smallest possible value. 17th Chinese 2002 © John Scholes jscholes@kalva.demon.co.uk 30 Sep 2002