Triangle sides

Knowing that $\max(x,y,z) \ge \dfrac{x+y+z}3 \Rightarrow \min( \max(x,y,z)) \ge \dfrac{\min(x+y+z)}3$ . Let $x = a+b,y=a+c, z=b+c$ , we have

$\max(a+b,a+c,b+c) \ge \dfrac{(a+b)+(b+c)+(c+a)}3 = \dfrac{24\times2}3 = 16$

Hence the minimum value of the maximum values of the numbers $a+b,a+c,b+c$ is $\boxed{16}$ and it can be achieved when $a+b=a+c=b+c$ or $a=b=c=8$ .