Define the sequence sequence $a_0,a_1, a_2,....,a_{2018}, a_{2019}$ of real numbers as follows: $\bullet$ $a_0 = 1$. $\bullet$ $a_{n + 1} = a_n - \frac{a_n^2}{2019}$ for $n = 0, 1, ...,2018$. Prove that $a_{2019} < \frac12 <a_{2018}$.
Source: Peru EGMO TST 2019 p5
Tags: recurrence relation, Sequence, inequalities, algebra
Define the sequence sequence $a_0,a_1, a_2,....,a_{2018}, a_{2019}$ of real numbers as follows: $\bullet$ $a_0 = 1$. $\bullet$ $a_{n + 1} = a_n - \frac{a_n^2}{2019}$ for $n = 0, 1, ...,2018$. Prove that $a_{2019} < \frac12 <a_{2018}$.