IMO 1993 SL 26

Origin: VIE

Problem

Let a, b, c, d be four nonnegative numbers satisfying a+b+c+d =

Solution

Set f(a, b, c, d) = abc + bcd + cda + dab −176 27 abcd = ab(c + d) + cd  a + b −176 27 ab  . If a + b −176 a b \leq0, by the arithmetic-geometric inequality we have f(a, b, c, d) \leqab(c + d) \leq 27. On the other hand, if a + b −176 a b > 0, the value of f increases if c, d are replaced by c+d 2 , c+d 2 . Consider now the following fourtuplets: P0(a, b, c, d), P1  a, b, c + d , c + d  , P2 a + b , a + b , c + d , c + d  , P3 1 4, a + b , c + d , 1  , P4 1 4, 1 4, 1 4, 1  From the above considerations we deduce that for i = 0, 1, 2, 3 either f(Pi) \leqf(Pi+1), or directly f(Pi) \leq1/27. Since f(P4) = 1/27, in every case we are led to f(a, b, c, d) = f(P0) \leq1 27. Equality occurs only in the cases (0, 1/3, 1/3, 1/3) (with permutations) and (1/4, 1/4, 1/4, 1/4). Remark. Lagrange multipliers also work. On the boundary of the set one of the numbers a, b, c, d is 0, and the inequality immediately follows, while for an extremum point in the interior, among a, b, c, d there are at most two distinct values, in which case one easily verifies the inequality.