IMO 2003 SL A4

Origin: IRE | Category: Algebra

Problem

Let n be a positive integer and let x1 \leqx2 \leq\cdot \cdot \cdot \leqxn be real numbers. (a) Prove that ⎛ ⎝ n  i,j=1 |xi −xj| ⎞ ⎠ \leq2(n2 −1) n  i,j=1 (xi −xj)2. (b) Show that equality holds if and only if x1, . . . , xn is an arithmetic progession.

Solution

By the Cauchy–Schwarz inequality we have ⎛ ⎝ n  i,j=1 (i −j)2 ⎞ ⎠ ⎛ ⎝ n  i,j=1 (xi −xj)2 ⎞ ⎠\geq ⎛ ⎝ n  i,j=1 |i −j| \cdot |xi −xj| ⎞ ⎠ . (1) On the other hand, it is easy to prove (for example by induction) that n  i,j=1 (i −j)2 = (2n −2) \cdot 12 + (2n −4) \cdot 22 + \cdot \cdot \cdot + 2 \cdot (n −1)2 = n2(n2 −1) and that n  i,j=1 |i −j| \cdot |xi −xj| = n n  i,j=1 |xi −xj|.

Thus the inequality (1) becomes n2(n2 −1) ⎛ ⎝ n  i,j=1 (xi −xj)2 ⎞ ⎠\geqn2 ⎛ ⎝ n  i,j=1 |xi −xj| ⎞ ⎠ , which is equivalent to the required one.