IMO 1991 SL 26

Origin: CZS

Problem

Let n \geq2 be a natural number and let the real numbers p, a1, a2, . . . , an, b1, b2, . . . , bn satisfy 1/2 \leqp \leq1, 0 \leqai, 0 \leqbi \leqp, i = 1, . . . , n, and n i=1 ai = n i=1 bi = 1. Prove the inequality n  i=1 bi n

j=1 j̸=i aj \leq p (n −1)n−1 .

Solution

Without loss of generality we can assume b1 \geqb2 \geq\cdot \cdot \cdot \geqbn. We denote by Ai the product a1a2 . . . ai−1ai+1 . . . an. If for some i < j holds Ai < Aj, then biAi + bjAj \leqbiAj + bjAi (or equivalently (bi −bj)(Ai −Aj) \leq0). Therefore the sum n i=1 biAi does not decrease when we rearrange the numbers a1, . . . , an so that A1 \geq\cdot \cdot \cdot \geqAn, and consequently a1 \leq\cdot \cdot \cdot \leq an. Further, for fixed ai’s and  bi = 1, the sum n i=1 biAi is maximal when b1 takes the largest possible value, i.e. b1 = p, b2 takes the remaining largest possible value b2 = 1 −p, whereas b3 = \cdot \cdot \cdot = bn = 0. In this case n  i=1 biAi = pA1 + (1 −p)A2 = a3 . . . an(pa2 + (1 −p)a1) \leqp(a1 + a2)a3 . . . an \leq p (n −1)n−1 , using the inequality between the geometric and arithmetic means for a3, . . . , an, a1 + a2.