IMO Shortlist 1996 problem A3


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2. travnja 2012.
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Let a > 2 be given, and starting a_0 = 1, a_1 = a define recursively:

a_{n+1} = \left(\frac{a^2_n}{a^2_{n-1}} - 2 \right) \cdot a_n.

Show that for all integers k > 0, we have: \sum^k_{i = 0} \frac{1}{a_i} < \frac12 \cdot (2 + a - \sqrt{a^2-4}).
Izvor: Međunarodna matematička olimpijada, shortlist 1996