equality holds for n even and half the x i 2, and half 0. Solution wlog x 1 ≥ x 2 ≥ ... ≥ x n . Then lhs = 2 ∑ i<j (x i -x j ) = 2(n-1)x 1 + 2(n-3)x 2 + ... - 2(n-1)x n . Suppose n = 2m. Then for i = 1, 2, ... , m the coefficient of x i is positive, so to maximise the sum we should take x 1 = x 2 = ... = x m = 2. Similarly for i = m+1, ... , n the coefficient of x i is negative, so to maximise the sum we should take x m+1 = x m+2 = ... = x n = 0. Then we get the maximum value 4(2m-1 + 2m-3 + ... + 1) = n 2 . Similarly, if n = 2m+1, then for i = 1, 2, ... , m the coefficient of x i is positive, so to maximise the sum we should take x 1 = x 2 = ... = x m = 2. The coefficient of xm+1 is zero, so its value is irrelevant. The coefficients of x m+2 , ... , x n are negative, so to maximise the sum we should take x m+2 = x m+2 = ... = x n = 0. Then we get the maximum value 4(2m + 2m-2 + ... + 2) = 4m(m+1) = (n-1)(n+1) = n 2 - 1. So for n odd the inequality still holds, but we cannot get equality. 21st BMO 1985 © John Scholes [email protected] 11 February 2004 Last corrected/updated 11 Feb 04