IMO 1995 SL A2

Origin: SWE | Category: Algebra

Problem

Let a and b be nonnegative integers such that ab \geqc2, where c is an integer. Prove that there is a number n and integers x1, x2, . . . , xn, y1, y2, . . . , yn such that n  i=1 x2 i = a, n  i=1 y2 i = b, and n  i=1 xiyi = c.

Solution

From the Cauchy–Schwarz inequality, [(y + z) + (z + x) + (x + y)] \cdot S \geq(x + y + z)2 ⇒ S \geqx + y + z . It follows from the A-G mean inequality that x+y+z \geq3 3\sqrtxyz = 3 2; hence the proof is complete. Equality holds if and only if x = y = z = 1, i.e., a = b = c = 1. Remark. After reducing the problem to x2 y+z + y2 z+x + z2 x+y \geq3 2, we can solve the problem using Jensen’s inequality applied to the function g(u, v) = u2/v. The problem can also be solved using Muirhead’s inequality. 2. We may assume c \geq0 (otherwise, we may simply put −yi in the place of yi). Also, we may assume a \geqb. If b \geqc, it is enough to take n = a+b−c, x1 = \cdot \cdot \cdot = xa = 1, y1 = \cdot \cdot \cdot = yc = ya+1 = \cdot \cdot \cdot = ya+b−c = 1, and the other xi’s and yi’s equal to 0, so we need only consider the case a > c > b. We proceed to prove the statement of the problem by induction on a + b. The case a+b = 1 is trivial. Assume that the statement is true when a+b \leq N, and let a+b = N+1. The triple (a+b−2c, b, c−b) satisfies the condition (since (a + b −2c)b −(c −b)2 = ab −c2), so by the induction hypothesis there are n-tuples (xi)n i=1 and (yi)n i=1 with the wanted property. It is easy to verify that (xi + yi)n i=1 and (yi)n i=1 give a solution for (a, b, c).