IMO 1995 SL A6

Origin: JAP | Category: Algebra

Problem

Let n be an integer, n \geq3. Let x1, x2, . . . , xn be real numbers such that xi < xi+1 for 1 \leqi \leqn −1. Prove that n(n −1)  i n−1  i=1 (n −i)xi ⎛ ⎝ n  j=2 (j −1)xj ⎞ ⎠.

Solution

Let yi = xi+1 + \cdot \cdot \cdot + xn, Y = n j=2(j −1)xj, and zi = n(n−1) yi −(n − i)Y . Then n(n−1)  i<j xixj − n−1 i=1 (n −i)xi

Y = n(n−1) n−1 i=1 xiyi − n−1 i=1 (n−i)xiY = n−1 i=1 xizi, so it remains to show that n−1 i=1 xizi > 0. Since n−1 i=1 yi = Y and n−1 i=1 (n −i) = n(n−1) , we have  zi = 0. Note that Y < n j=2(j −1)xn = n(n−1) xn, and consequently zn−1 = n(n−1) xn −Y > 0. Furthermore, we have zi+1 n −i −1 − zi n −i = n(n −1)  yi+1 n −i −1 − yi n −i 

0, which means that z1 n−1 < z2 n−2 < \cdot \cdot \cdot < zn−1 1 . Therefore there is a k for which z1, . . . , zk \leq0 and zk+1, . . . , zn−1 > 0. But then zi(xi −xk) \geq0, i.e., xizi \geqxkzi for all i, so n−1 i=1 xizi > n−1 i=1 xkzi = 0 as required. Second solution. Set X = n−1 j=1 (n −j)xj and Y = n j=2(j −1)xj. Since 4XY = (X + Y )2 −(X −Y )2, the RHS of the inequality becomes XY = 1 ⎡ ⎣(n −1)2

n  i=1 xi −

n  i=1 (2i −1 −n)xi 2⎤ ⎦. The LHS equals 1  (n −1)2 (n i=1 xi)2 −(n −1)  i<j(xj −xi)2 . Since n i=1(2i −1 −n)xi =  i<j(xj −xi) also holds, we must prove that ⎛ ⎝ i<j (xj −xi) ⎞ ⎠

(n −1)  i<j (xj −xi)2. (1) Putting xi+1 −xi = di > 0 (so, xj −xi = di + di+1 + \cdot \cdot \cdot + dj−1) and expanding the obtained expressions, we reduce this inequality to  k k2(n −k)2d2 k + 2  k<l kl(n −k)(n −l)dkdl >  k(n −1)k(n −k)d2 k + 2  k<l(n −1)k(n −l)dkdl, which is verified immediately by comparing coefficients. Remark. An inequality significantly stronger than (1) in the second solu- tion has appeared later, as IMO 03-5.