Consider a sequence of real numbers defined by: $a_{n + 1} = a_n + \frac{1}{a_n}$ for $n = 0, 1, 2, ...$ Prove that, for any positive real number $a_0$, is true that $a_{1996}$ is greater than $63$.
Problem
Source: Cono sur 1996, Problem 2
Tags: cono sur, Sequences
WolfusA
11.06.2018 19:30
Similar to 1975 IMO SL https://artofproblemsolving.com/community/c6h367919p2025382
Thelink_20
13.05.2024 22:00
$a_{n+1}=a_{n}+\frac{1}{a_{n}}\iff a_{n+1}^2=a_{n}^2+\frac{1}{a_{n}^2}+2\iff a_{n+1}^2-a_{n}^2=\frac{1}{a_{n}^2}+2$ $\begin{cases}a_{1996}^2-a_{1995}^2=\frac{1}{a_{1995}^2}+2 \\ a_{1995}^2-a_{1994}^2=\frac{1}{a_{1994}^2}+2 \\ \dots \\ a_{1}^2-a_{0}^2=\frac{1}{a_{0}^2}+2 \end{cases} \Rightarrow a_{1996}^2=a_{0}^2+\sum_{i=0}^{1995}\frac{1}{a_{i}^2}+2.1996\Rightarrow a_{1996}^2\geq 3992>63^2\Rightarrow a_{1996}>63$. And we are done.