Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Prove it

罒 (Full marks 20 points) Let the function $f(x)$ have the domain $\mathbf{R}$ . For any $x_{1}, x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$ , it holds that $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| < \left|x_{1}-x_{2}\right|$ . Given that there exists a real number $\alpha$ such that $f(\alpha)=\alpha$ . If the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}<\alpha, f\left(a_{n}\right)=2 a_{n+1}-a_{n}$ , prove that when $n \in \mathrm{N}$ , we have
(i) $a_{n}<\alpha$ ;
(ii) the sequence $\left\{a_{n}\right\}$ is an increasing sequence.

Solution

$\begin{array}{l} \text{(i) When } n=1, \because a_{1}a_{k} . \end{array}\right. \end{array}$

From (1) we get $2 a_{k+1}-a_{k}+a_{k}a_{n}$ , then $2 a_{n+1}-a_{n}>a_{n}$ . Therefore, $a_{n+1}>a_{n}$ , which means the sequence $\left\{a_{n}\right\}$ is an increasing sequence.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.