Team selection tests for IMO 2018

1) Put $a_1=b_1=x$ and $a_n=b_n=y$, then $0<x<y$. Denote $d$ as the difference and the ratio of two consecutive terms of $(a_n)$, $(b_n)$ respectively. We have $a_n=a_1+(n-1)d$ and $a_k=a_1+(k-1)d$ for any $k\in \{2,3,\dots ,n-1\}$. Hence, $$a_k=\frac{(n-k)a_1+(k-1)a_n}{n-1}=\frac{(n-k)x+(k-1)y}{n-1}.$$ Similarly, $b_k=\sqrt[n-1]{x^{n-k}{y}^{k-1}}$ so we need to prove that $$\frac{(n-k)x+(k-1)y}{n-1}>\sqrt[n-1]{x^{n-k}{y}^{k-1}},$$ which is true by AM-GM inequality.

2) With $2^k\le m$, consider some estimations as follow $${\left(1+\frac{1}{m}\right)}^k=1+\frac{\left(\genfrac{}{}{0ex}{}{k}{1}\right)}{m}+\frac{\left(\genfrac{}{}{0ex}{}{k}{2}\right)}{m^2}+\cdots \ge 1+\frac{k}{m}$$ and $$\begin{array}{rl}{\left(1+\frac{1}{m}\right)}^k& =1+\frac{\left(\genfrac{}{}{0ex}{}{k}{1}\right)}{m}+\frac{\left(\genfrac{}{}{0ex}{}{k}{2}\right)}{m^2}+\cdots \le 1+\frac{\left(\genfrac{}{}{0ex}{}{k}{1}\right)}{m}+\frac{\left(\genfrac{}{}{0ex}{}{k}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{k}{k}\right)}{m}\\ & <1+\frac{k}{m}+\frac{2^k}{m^2}\le 1+\frac{k+1}{m}\end{array}$$ then $1+\frac{k}{m}<{\left(1+\frac{1}{m}\right)}^k<1+\frac{k+1}{m}$ or $m+k<m{\left(1+\frac{1}{m}\right)}^k<m+k+1$. This is true for all $k\le {\log }_{2}m$ so $$m\left(1+\frac{1}{m}\right)<m+2<m{\left(1+\frac{1}{m}\right)}^2<m+3<\cdots <m+k+1.$$ To make all these numbers are integers, we multiply them by $m^{[{\log }_{2}m]}$. Finally, take ${\log }_{2}m>n$, the result will follow. $\square$