IMO 1998 SL 10

Origin: AUS

Problem

Let r1, r2, . . . , rn be real numbers greater than or equal to 1. Prove that r1 + 1 + r2 + 1 + \cdot \cdot \cdot + rn + 1 \geq n n\sqrtr1r2 \cdot \cdot \cdot rn + 1.

Solution

We shall first prove the inequality for n of the form 2k, k = 0, 1, 2, . . .. The case k = 0 is clear. For k = 1, we have r1 + 1 + r2 + 1 − \sqrtr1r2 + 1 = (\sqrtr1r2 −1)(\sqrtr1 −\sqrtr2)2 (r1 + 1)(r2 + 1)(\sqrtr1r2 + 1) \geq0. For the inductive step it suffices to show that the claim for k and 2 implies that for k + 1. Indeed, 2k+1  i=1 ri + 1 \geq 2k 2k\sqrtr1r2 \cdot \cdot \cdot r2k + 1 + 2k 2k\sqrtr2k+1r2k+2 \cdot \cdot \cdot r2k+1 + 1 \geq 2k+1 2k+1\sqrtr1r2 \cdot \cdot \cdot r2k+1 + 1 , (1) and the induction is complete. We now show that if the statement holds for 2k, then it holds for every n < 2k as well. Put rn+1 = rn+2 = \cdot \cdot \cdot = r2k = n\sqrtr1r2 . . . rn. Then (1) becomes r1 + 1 + \cdot \cdot \cdot + rn + 1 + 2k −n n\sqrtr1 \cdot \cdot \cdot rn + 1 \geq 2k n\sqrtr1 \cdot \cdot \cdot rn + 1. This proves the claim. Second solution. Define ri = exi, where xi > 0. The function f(x) = 1+ex is convex for x > 0: indeed, f ′′(x) = ex(ex−1) (ex+1)3 > 0. Thus by Jensen’s in- equality applied to f(x1), . . . , f(xn), we get r1+1 +\cdot \cdot \cdot+ rn+1 \geq n n\sqrtr1\cdot\cdot\cdotrn+1.