Determine all possible values of a/(a+b+d) + b/(a+b+c) + c/(b+c+d) + d/(a+c+d) for positive reals a, b, c, d.
Solution
We show first that the sum must lie between 1 and 2. If we replace each denominator by a+b+c+d then we reduce each term and get 1. Hence the sum is more than 1. Suppose a is the largest of the four reals. Then the first term is less than 1. The second and fourth terms have denominators greater than b+c+d, so the terms are increased if we replace the denominators by b+c+d. But then the last three terms sum to 1. Thus the sum of the last three terms is less than 1. Hence the sum is less than 2.
If we set a = c = 1 and make b and d small, then the first and third terms can be made arbitarily close to 1 and the other two terms arbitarily close to 0, so we can make the sum arbitarily close to 2. If we set a = 1, c = d and make b and c/b arbitarily small, then the first term is arbitarily close to 1 and the last three terms are all arbitarily small, so we can make the sum arbitarily close to 1. Hence, by continuity, we can achieve any value in the open interval (1,2).
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
7 Oct 1998