7th Iberoamerican 1992 Problem A2 Let f(x) = a 1 /(x + a 1 ) + a 2 /(x + a 2 ) + ... + a n /(x + a n ), where a i are unequal positive reals. Find the sum of the lengths of the intervals in which f(x) ≥ 1.
∑ a i Solution wlog a 1 > a 2 > ... > a n . The graph of each a i /(x + a i ) is a rectangular hyberbola with asymptotes x = -a i and y = 0. So it is not hard to see that the graph of f(x) is made up of n+1 strictly decreasing parts. For x  -a n , f(x) decreases from ∞ to 0. Thus f(x) = 1 at n values b 1 < b 2 < ... < b n , and f(x) ≥ 1 on the n intervals (-a 1 ,b 1 ), (-a 2 ,b 2 ), ... , (-a n ,b n ). So the sum of the lengths of these intervals is ∑ (a i + b i ). We show that ∑ b i = 0. Multiplying f(x) = 1 by ∏(x + a j ) we get a polynomial of degree n: ∏(x + a j ) - ∑ i (a i ∏ j≠i (x + a j ) ) = 0 . The coefficient of x n is 1 and the coefficient of x n-1 is ∑ a j - ∑ a i = 0. Hence the sum of the roots, which is ∑ b i , is zero. Thanks to Juan Ignacio Restrepo 7th Ibero 1992 © John Scholes jscholes@kalva.demon.co.uk 29 March 2004