Dota skaitļu virkne \(x_{0}, x_{1}, x_{2}, x_{3}, \ldots\), kurā \(x_{0}>0\) un \(x_{n+1}=x_{n}+\frac{2}{x_{n}}\) visiem \(n \geq 0\). Pierādīt, ka \(x_{100}>20\).
Tā kā \(x_{0}>0\) un \(x_{n+1}=x_{n}+\frac{2}{x_{n}}\) katram \(n \geq 0\), tad \(x_{n}>0\) visiem \(n \geq 0\).
Aplūkosim skaitļu virkni \(y_{n}=x^{2}_{n}\) katram \(n \geq 0\).
Tad \(y_{n+1}=x_{n+1}^{2}=\left(x_{n}+\frac{2}{x_{n}}\right)^{2}=x_{n}^{2}+4+\frac{4}{x^{2}_{n}}>x_{n}^{2}+4=y_{n}+4\).
Tātad \(y_{100}>y_{0}+4 \cdot 100=y_{0}+400\) jeb \(x_{100}^{2}>x_{0}^{2}+400>0+400=400\). Tā kā \(x_{100}>0\) un \(x_{100}^{2}>400\), tad \(x_{100}>20\), k.b.j.