IMO 1999 SL A1

Origin: POL | Category: Algebra

Problem

Let n \geq2 be a fixed integer. Find the least constant C such that the inequality  i<j xixj(x2 i + x2 j) \leqC  i xi holds for every x1, . . . , xn \geq0 (the sum on the left consists of
n  sum- mands). For this constant C, characterize the instances of equality.

Solution

For all xi = 0 any C will do, so we may assume the contrary. Since the equation is symmetric and homogeneous, we may assume  i xi = 1. The equation now becomes F(x1, x2, . . . , xn) =  i<j xixj(x2 i + x2 j) =  i x2 i  j̸=i xj =  i x3 i (1 −xi) =  i f(xi) \leqC, where we define f(x) = x3 −x4. We note that for x, y \geq0 and x + y \leq2/3, f(x + y) + f(0) −f(x) −f(y) = 3xy(x + y) 2 3 −x −y  \geq0 . (1) We note that if at least three elements of {x1, x2, . . . , xn} are nonzero the condition of (1) always holds for the two smallest ones. Hence, applying (1) repeatedly, we obtain F(x1, x2, . . . , xn) \leqF(a, 1 −a, 0, . . . , 0) = 1 2(2a(1 − a))(1 −2a(1 −a)) \leq1 8 = F
1 2, 1 2, 0, . . . , 0  . Thus we have C = 1 8 (for all

n), and equality holds only when two xi are equal and the remaining ones are 0. Second solution. Let M = x2 1 + x2 2 + \cdot \cdot \cdot + x2 n. Using ab \leq(a + 2b)2/8 we have  1\leqi<j\leqn xixj(x2 i + x2 j) \leqM  i<j xixj \leq1 ⎛ ⎝M + 2  i<j xixj ⎞ ⎠ = 1

n  i=1 xi . Equality holds if and only if M = 2  i<j xixj and xixj(x2 i +x2 j) = Mxixj for all i < j, which holds if and only if n −2 of the xi are zero and the remaining two are equal. Remark. Problems (SL90-26) and (SL91-27) are very similar.