Team Selection Test for JBMO 2024

Answer: 2460. Since $$\frac{a_{n+1}}{b_{n+1}}=\frac{a_n+\frac{1}{b_n}}{b_n+\frac{1}{a_n}}=\frac{a_n}{b_n}$$ we get that $\frac{a_n}{b_n}$ is a constant and hence equals $\frac{40}{41}$. Therefore, $a_k>80$ is equivalent to $b_k>82$ and $a_k{b}_k>80\cdot 82=6560$. Multiplying both sequences we get $$a_{n+1}{b}_{n+1}=a_n{b}_n+2+\frac{1}{a_n{b}_n}$$ Hence, for any $k$ we have $a_k{b}_k>40\cdot 41+2k$ which means $a_k{b}_k>6560$ for $k=2460$. Moreover, if $k=2459$, we have $$a_k{b}_k=6560-2+\sum _{i=0}^{k-1}\frac{1}{a_i{b}_i}<6560-2+\frac{2459}{1640}<6560.$$ Thus, the minimal value of $k$ satisfying conditions is 2460.