, 2015

$\boxed{30}$

Let $g_1 = \text{gcd}\{a, b+c\}$, $g_2 = \text{gcd}\{b, c+a\}$, $g_3 = \text{gcd}\{c, a+b\}$. If there exists a prime $p$ which divides both $g_1$ and $g_2$, then $p$ must divide both $a$ and $b$. Furthermore, since $p$ divides $b+c$ as well, $c$ must be divisible by $p$, but this contradicts the assumption that $\text{gcd}\{a, b, c\} = 1$. Therefore, we conclude that $g_1$ and $g_2$ are relatively prime. Similarly, $g_2$ and $g_3$ are relatively prime and so are $g_3$ and $g_1$. Combining these facts with the assumption that $g_1$, $g_2$, $g_3$ are all greater than $1$, we get $g_1g_2g_3 \ge 2 \cdot 3 \cdot 5 = 30$.

Furthermore, since $g_1$ divides both $a$ and $b+c$, it divides $a+b+c$ as well. Similarly, both $g_2$ and $g_3$ divide $a+b+c$. Therefore, we conclude that the product $g_1g_2g_3$ divides $a+b+c$ also, and this implies that $a+b+c \ge g_1g_2g_3 \ge 30$ must hold.

On the other hand, we see that the triple $(a, b, c) = (2, 3, 25)$ satisfies all the conditions of the problem and that $a + b + c = 30$ is satisfied. Therefore, we conclude that $30$ is the minimum value we seek.