IMO 1982 SL 3

Problem

A3 (USS 4)IMO3 Consider the infinite sequences {xn} of positive real numbers with the following properties: x0 = 1 and for all i \geq0, xi+1 \leqxi. (a) Prove that for every such sequence there is an n \geq1 such that x2 x1 + x2 x2 + \cdot \cdot \cdot + x2 n−1 xn \geq3.999. (b) Find such a sequence for which x2 x1 + x2 x2 + \cdot \cdot \cdot + x2 n−1 xn < 4 for all n.

Solution

(a) By the Cauchy–Schwarz inequality we have
x2 0/x1 + \cdot \cdot \cdot + x2 n−1/xn  \cdot (x1 + \cdot \cdot \cdot + xn) \geq(x0 + \cdot \cdot \cdot + xn−1)2. Let us set Xn−1 = x1 + x2 + \cdot \cdot \cdot + xn−1. Using x0 = 1, the last inequality can be rewritten as x2 x1

• \cdot \cdot \cdot + x2 n−1 xn \geq(1 + Xn−1)2 Xn−1 + xn \geq 4Xn−1 Xn−1 + xn = 1 + xn/Xn−1 . (1) Since xn \leqxn−1 \leq\cdot \cdot \cdot \leqx1, it follows that Xn−1 \geq(n −1)xn. Now (1) yields x2 0/x1 + \cdot \cdot \cdot + x2 n−1/xn \geq4(n −1)/n, which exceeds 3.999 for n > 4000. (b) The sequence xn = 1/2n obviously satisfies the required condition. Second solution to part (a). For each n \inN, let us find a constant cn such that the inequality x2 0/x1 + \cdot \cdot \cdot + x2 n−1/xn \geqcnx0 holds for any sequence x0 \geqx1 \geq\cdot \cdot \cdot \geqxn > 0. For n = 1 we can take c1 = 1. Assuming that cn exists, we have x2 x1

x2 x2

• \cdot \cdot \cdot + x2 n xn+1  \geqx2 x1
• cnx1 \geq2

x2 0cn = x0 \cdot 2\sqrtcn. Thus we can take cn+1 = 2\sqrtcn. Then inductively cn = 22−1/2n−2, and since cn \to4 as n \to\infty, the result follows. Third solution. Since {xn} is decreasing, there exists limn\to\inftyxn = x \geq0. If x > 0, then x2 n−1/xn \geqxn \geqx holds for each n, and the result is trivial.

If otherwise x = 0, then we note that x2 n−1/xn \geq4(xn−1 −xn) for each n, with equality if and only if xn−1 = 2xn. Hence lim n\to\infty n  k=1 x2 k−1 xk \geqlim n\to\infty n  k=1 4(xk−1 −xk) = 4x0 = 4. Equality holds if and only if xn−1 = 2xn for all n, and consequently xn = 1/2n.