Answer

The minimum possible value of $a_1 + a_2 + a_3$ is $38$. Bound: $a_1 + a_2 + a_3 \geq 38$ First, we claim that $a_1 a_2 a_3 \geq 2025$. Assume otherwise. Each vertex of the graph has $a_1$ choices for the first partition, $a_2$ for the second, and $a_3$ for the third. Thus, there are a total of $a_1 a_2 a_3$ ways to assign a vertex across all three partitions. Since we assumed $a_1 a_2 a_3 < 2025$, the Pigeonhole Principle implies that there must exist two vertices $u$ and $v$ that are in the same group for all three partitions. However, this means the edge $uv$ is not colored with $r_1$, $r_2$, or $r_3$, which is a contradiction. To finish, we apply AM-GM: \[ a_1 + a_2 + a_3 \geq \lceil 3\sqrt[3]{a_1 a_2 a_3} \rceil \geq \lceil 3\sqrt[3]{2025} \rceil \geq \lceil 30\sqrt[3]{2} \rceil \geq \lceil 30 \cdot \frac{5}{4} \rceil = 38. \] Construction: $a_1 + a_2 + a_3 = 38$ Consider the $12 \times 13 \times 13 > 2025$ lattice points $(x, y, z)$ with $1 \leq x \leq 12$ and $1 \leq y, z \leq 13$. Arbitrarily assign the 2025 vertices to these coordinates. Any two vertices $u$ and $v$ must differ in at least one coordinate, say in position $i$. Then assign the color $r_i$ to the edge connecting $u$ and $v$. For each $i = 1, 2, 3$, group the vertices based on the value in position $i$. Since two vertices sharing the same value in position $i$ would not have an edge colored with $r_i$, this construction is valid. Thus, we achieved $a_1 = 12$, $a_2 = 13$, and $a_3 = 13$, yielding $a_1 + a_2 + a_3 = 38$, as desired.