IMO 1982 SL 15

Problem

C3 (CAN 5) Show that 1 −sa 1 −s \leq(1 + s)a−1 holds for every 1 ̸= s > 0 real and 0 < a \leq1 rational.

Solution

Let a = k/n, where n, k \inN, n \geqk. Putting tn = s, the given inequality becomes 1−tk 1−tn \leq(1 + tn)k/n−1, or equivalently (1 + t + \cdot \cdot \cdot + tk−1)n(1 + tn)n−k \leq(1 + t + \cdot \cdot \cdot + tn−1)n. This is clearly true for k = n. Therefore it is enough to prove that the left- hand side of the above inequality is an increasing function of k. We are led to show that (1+t+\cdot \cdot \cdot+tk−1)n(1+tn)n−k \leq(1+t+\cdot \cdot \cdot+tk)n(1+tn)n−k−1. This is equivalent to 1 + tn \leqAn, where A = 1+t+\cdot\cdot\cdot+tk 1+t+\cdot\cdot\cdot+tk−1 . But this easily follows, since An −tn = (A −t)(An−1 + An−2t + \cdot \cdot \cdot + tn−1) \geq(A −t)(1 + t + \cdot \cdot \cdot + tn−1) = 1 + t + \cdot \cdot \cdot + tn−1 1 + t + \cdot \cdot \cdot + tk−1 \geq1. Remark. The original problem asked to prove the inequality for real a.