IMO 1995 SL 26

Problem

S4 (NZL) Suppose that x1, x2, x3, . . . are positive real numbers for which xn n = n−1  j=0 xj n for n = 1, 2, 3, . . .. Prove that for all n, 2 − 2n−1 \leqxn < 2 −1 2n .

Solution

For n = 1 the result is trivial, since x1 = 1. Suppose now that n \geq2 and let fn(x) = xn −n−1 i=0 xi. Note that xn is the unique positive real root of fn, because fn(x) xn−1 = x −1 −1 x −\cdot \cdot \cdot − xn−1 is strictly increasing on R+. Consider gn(x) = (x −1)fn(x) = (x −2)xn + 1. Obviously gn(x) has no positive roots other than 1 and xn > 1. Observe that
1 − 2n n > 1 −n 2n \geq1 2 for n \geq2 (by Bernoulli’s inequality). Since then gn  2 −1 2n  = −1 2n  2 −1 2n n

• 1 = 1 −  1 − 2n+1 n

0, and gn  2 − 2n−1  = − 2n−1  2 − 2n−1 n

• 1 = 1 −2  1 −1 2n n < 0, we conclude that xn is between 2 − 2n−1 and 2 − 2n , as required. Remark. Moreover, limn\to\infty2n(2 −xn) = 1.