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Inequality on recurrence

Source: 8th European Mathematical Cup, Junior Category, Q2

December 26, 2019
inequalitiesalgebra

Problem Statement

Let (xn)n∈N(x_n)_{n\in \mathbb{N}} be a sequence defined recursively such that x1=2x_1=\sqrt{2} and xn+1=xn+1xn for n∈N.x_{n+1}=x_n+\frac{1}{x_n}\text{ for }n\in \mathbb{N}. Prove that the following inequality holds: x122x1x2−1+x222x2x3−1+…+x201822x2018x2019−1+x201922x2019x2020−1>20192x20192+1x20192.\frac{x_1^2}{2x_1x_2-1}+\frac{x_2^2}{2x_2x_3-1}+\dotsc +\frac{x_{2018}^2}{2x_{2018}x_{2019}-1}+\frac{x_{2019}^2}{2x_{2019}x_{2020}-1}>\frac{2019^2}{x_{2019}^2+\frac{1}{x_{2019}^2}}.
Proposed by Ivan Novak
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