Library / Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads /33 of 80

Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Algebra Difficulty 8.1 Shortlist Prove it Romania

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

Solution

Use the relation in the statement and the fact that no $a_k$ is $1/2$ , to write
$b_{k+2}^2 - b_k^2 = \frac{(b_{k+1} - b_k)^2 - (b_{k+2} - b_{k+1})^2}{2a_k - 1}, \quad k \in \mathbb{N},$
fix an integer $n > 2$ , and sum over $0, 1, \ldots, n-2$ , to obtain
$b_n^2 + b_{n-1}^2 - b_1^2 - b_0^2 = \frac{(b_1 - b_0)^2}{2a_0 - 1} - \sum_{k=0}^{n-3} \left( \frac{1}{2a_k - 1} - \frac{1}{2a_{k+1} - 1} \right) (b_{k+2} - b_{k+1})^2 - \frac{(b_n - b_{n-1})^2}{2a_{n-2} - 1} \le \frac{(b_1 - b_0)^2}{2a_0 - 1},$
since the $a_k$ form an increasing sequence of real numbers greater than $1/2$ . Consequently, $b_n^2 \le b_n^2 + b_{n-1}^2 \le b_0^2 + b_1^2 + (b_1 - b_0)^2/(2a_0 - 1)$ , and the conclusion follows.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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