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Algebra Difficulty 6.4 National olympiad Prove it Belarus

f(x)={x+12,if x<12,x2,if x12. f(x) = \begin{cases} x + \frac{1}{2}, & \text{if } x < \frac{1}{2}, \\ x^2, & \text{if } x \ge \frac{1}{2}. \end{cases}
Let aa and bb be two real numbers such that 0<a<b<10 < a < b < 1. We define the sequences ana_n and bnb_n by a0=a,b0=ba_0 = a, b_0 = b, and
an=f(an1), bn=f(bn1) for n>0. a_n = f(a_{n-1}),\ b_n = f(b_{n-1}) \text{ for } n > 0.
Show that there exists a positive integer nn such that
(anan1)(bnbn1)<0. (a_n - a_{n-1})(b_n - b_{n-1}) < 0.

Solution

2. See IMO-2014 Shortlist, Problem A2.

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