Any three small triangles, from which no two have common vertices, take up nine of the ten numbers written into the vertices of the small triangles. So, the sum of the numbers inside those small triangles is $55 - a$, where $a$ is the number at the last vertex. Now it is sufficient to prove that we can choose the three small triangles, from which no two have common vertices, in four different ways, always leaving a different vertex out. So, in at least one case, the number at the last vertex is at most $7$, and the sum of the numbers in the three chosen triangles is at least $55 - 7 = 48$.

Fig. 11

Fig. 12

Fig. 13

Fig. 14

Indeed, the three triangles can be chosen so that they leave uncovered the central number (Fig. 11) or one of the corner numbers (Fig. 12, 13, 14).

Let $m$ be the number in the center of the large triangle. Then, when adding the sums of the three corner triangles we get the sum of all the numbers from $1$ to $10$, except $m$, so the sum of the corner triangles is $55 - m$. If $m \le 7$, then the sum is at least $48$.

In the rest of the cases, consider any three triangles around the center point, such that no two of them share a side. Adding the numbers in them, we get the sum of all the numbers from $1$ to $10$, except the three numbers in the corners, while we add the center number three times. So the sum of those triangles is at least $21 + 3m$. If $m \ge 9$, then the sum is at least $48$.

This leaves the case $m = 8$. The sum of the numbers in the triangles in the corners is at least $55 - 8 = 47$, so at least one of them contains a sum that is at least $16$. If none of the triangles contains a sum $17$ or greater, the number $16$ must occur in two different triangles. The sum of the numbers in any three triangles around the center point, chosen like above, is at least $21 + 3 \cdot 8 = 45$. So, in both triples at least one of the triangles contains a sum of at least $15$, and since triangles sharing an edge cannot contain the same sum as the corresponding sums differ by exactly one term, at least one of the six triangles around the center point contains a sum of at least $16$. Thus we can pick the three desired triangles from among either one corner triangle and two central triangles or two corner triangle and one central triangle that contain the largest numbers.