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Algebra Difficulty 8.1 Shortlist Prove it Romania

Let (an)n≥0(a_n)_{n \ge 0} and (bn)n≥0(b_n)_{n \ge 0} be sequences of real numbers such that a0>1/2a_0 > 1/2, an+1≥ana_{n+1} \ge a_n, and bn+1=an(bn+bn+2)b_{n+1} = a_n(b_n + b_{n+2}), for all non-negative integers nn. Show that the sequence (bn)n≥0(b_n)_{n \ge 0} is bounded.

Solution

Use the relation in the statement and the fact that no aka_k is 1/21/2, to write
bk+22−bk2=(bk+1−bk)2−(bk+2−bk+1)22ak−1,k∈N, b_{k+2}^2 - b_k^2 = \frac{(b_{k+1} - b_k)^2 - (b_{k+2} - b_{k+1})^2}{2a_k - 1}, \quad k \in \mathbb{N},
fix an integer n>2n > 2, and sum over 0,1,…,n−20, 1, \ldots, n-2, to obtain
bn2+bn−12−b12−b02=(b1−b0)22a0−1−∑k=0n−3(12ak−1−12ak+1−1)(bk+2−bk+1)2−(bn−bn−1)22an−2−1≤(b1−b0)22a0−1, b_n^2 + b_{n-1}^2 - b_1^2 - b_0^2 = \frac{(b_1 - b_0)^2}{2a_0 - 1} - \sum_{k=0}^{n-3} \left( \frac{1}{2a_k - 1} - \frac{1}{2a_{k+1} - 1} \right) (b_{k+2} - b_{k+1})^2 - \frac{(b_n - b_{n-1})^2}{2a_{n-2} - 1} \le \frac{(b_1 - b_0)^2}{2a_0 - 1},
since the aka_k form an increasing sequence of real numbers greater than 1/21/2. Consequently, bn2≤bn2+bn−12≤b02+b12+(b1−b0)2/(2a0−1)b_n^2 \le b_n^2 + b_{n-1}^2 \le b_0^2 + b_1^2 + (b_1 - b_0)^2/(2a_0 - 1), and the conclusion follows.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.