, 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 $k \in \{2, 3, \ldots, n-1\}$. 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 $2^{k} \leq m$, consider some estimations as follow
$$\left(1 + \frac{1}{m}\right)^{k} = 1 + \frac{\binom{k}{1}}{m} + \frac{\binom{k}{2}}{m^{2}} + \cdots \geq 1 + \frac{k}{m}$$
and
$$\begin{aligned} \left(1 + \frac{1}{m}\right)^{k} & = 1 + \frac{\binom{k}{1}}{m} + \frac{\binom{k}{2}}{m^{2}} + \cdots \leq 1 + \frac{\binom{k}{1}}{m} + \frac{\binom{k}{2} + \cdots + \binom{k}{k}}{m} \\ & < 1 + \frac{k}{m} + \frac{2^{k}}{m^{2}} \leq 1 + \frac{k+1}{m} \end{aligned}$$
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 $k \leq \log_{2} m$ 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. $\square$