IMO 1974 SL 8

Problem

II 2 (NET 3)IMO5 If a, b, c, d are arbitrary positive real numbers, find all possible values of S = a a + b + d + b a + b + c + c b + c + d + d a + c + d.

Solution

It is clear that a a + b + c + d + b a + b + c + d + c a + b + c + d + d a + b + c + d < S and S < a a + b + b a + b + c c + d + d c + d, or equivalently, 1 < S < 2. On the other hand, all values from (1, 2) are attained. Since S = 1 for (a, b, c, d) = (0, 0, 1, 1) and S = 2 for (a, b, c, d) = (0, 1, 0, 1), due to continuity all the values from (1, 2) are obtained, for example, for (a, b, c, d) = (x(1 −x), x, 1 −x, 1), where x goes through (0, 1). Second solution. Set S1 = a a + b + d + c b + c + d and S2 = b a + b + c + d a + c + d.

We may assume without loss of generality that a + b + c + d = 1. Putting a + c = x and b + d = y (then x + y = 1), we obtain that the set of values of S1 = a 1 −c + c 1 −a = 2ac + x −x2 ac + 1 −x is  x, 2x 2−x

. Having the analogous result for S2 in mind, we conclude that the values that S = S1 + S2 can take are  x + y, 2x 2−x + 2y 2−y  . Since x + y = 1 and 2x 2 −x + 2y 2 −y = 4 −4xy 2 + xy \leq2 with equality for xy = 0, the desired set of values for S is (1, 2).