, 2022

We assume that there exist $4m+2$ elements $x_{0} < x_{1} < \ldots < x_{4m+1}$ from $\{0,1,2, \ldots, 5^{m}\}$ for which the claim does not hold. Then in particular $x_{4m+1} + 2x_{i} \leq 3x_{i+1}$ holds for all $i = 0, 1, \ldots, 4m-1$. Rearranging gives $x_{4m+1} - x_{i} \geq \frac{3}{2}(x_{4m+1} - x_{i+1})$.
A simple induction argument yields from this $x_{4m+1} - x_{i} \geq \left(\frac{3}{2}\right)^{4m-i}(x_{4m+1} - x_{4m})$.
Here, setting $i=0$ leads to $x_{4m+1} - x_{0} \geq \left(\frac{3}{2}\right)^{4m}(x_{4m+1} - x_{4m}) = \left(\frac{81}{16}\right)^{m}(x_{4m+1} - x_{4m}) > 5^{m} \cdot 1$, a contradiction! ㅁ

We denote the largest element of $A$ by $c$. For $k = 0, \ldots, 4m-1$ we define $A_{k} = \{x \in A \mid (1 - (\frac{2}{3})^{k})c \leq x < (1 - (\frac{2}{3})^{k+1})c\}$.
Since $(1 - (\frac{2}{3})^{4m})c = c - (\frac{16}{81})^{m}c > c - (\frac{1}{5})^{m}c \geq c-1$, the sets $A_{0}, A_{1}, \ldots, A_{4m-1}$ form a partition of $A \setminus \{c\}$. Since $A \setminus \{c\}$ consists of $4m+1$ elements, by the pigeonhole principle there must exist a set $A_{k}$ that consists of at least two elements. We denote two of the numbers from $A_{k}$ by $a$ and $b$ such that $a < b < c$ holds. Then $c + 2a \geq c + 2(1 - (\frac{2}{3})^{k})c = (3 - 2(\frac{2}{3})^{k})c = 3(1 - (\frac{2}{3})^{k+1})c > 3b$, as required. $\square$