In a triangle, which has $10$ at one vertex, the sum is at least $13$. If $10$ is not at one of the vertices of the large triangle, the number of triangles with sum greater than $12$ is at least $3$ and the problem is solved. If $10$ is at the vertex of the large triangle, then look, where is the number $9$. If $9$ does not lie at a vertex of the large triangle, then there are at least $3$ small triangles, with a sum of at least $12$, and the problem is solved. If $9$ is at a vertex of the large triangle, then look, where is the number $8$. If $8$ does not lie at the vertex of a large triangle, then it is at a vertex of at least $3$ small triangles. In at least two of them the sum is at least $12$, and as at most one of these can overlap with one of the triangles found earlier, the problem is solved. If $8$ is at a vertex of the large triangle, then either the sum in that triangle is at least $12$, meaning the problem is solved, or the other vertices in the triangle have numbers $1$ and $2$. In the last case, the numbers $3$, $4$ and $5$ are the smallest numbers whose positions are not set, but these give a sum of $12$. So, we can simply choose any triangle whose vertices we have not looked at yet as our third triangle (there are three of these triangles).