IMO 1975 SL 4

Origin: SWE

Problem

Let a1, a2, . . . , an, . . . be a sequence of real numbers such that 0 \leqan \leq1 and an −2an+1 + an+2 \geq0 for n = 1, 2, 3, . . .. Prove that 0 \leq(n + 1)(an −an+1) \leq2 for n = 1, 2, 3, . . ..

Solution

Put ∆an = an −an+1. By the imposed condition, ∆an > ∆an+1. Suppose that for some n, ∆an < 0: Then for each k \geqn, ∆ak < ∆an; hence an −an+m = ∆an + \cdot \cdot \cdot + ∆an+m−1 < m∆an. Thus for sufficiently large m it holds that an −an+m < −1, which is impossible. This proves the first part of the inequality. Next one observes that n \geq n  k=1 ak = nan+1 + n  k=1 k∆ak \geq(1 + 2 + \cdot \cdot \cdot + n)∆an = n(n + 1) ∆an. Hence (n + 1)∆an \leq2.