We'll show that it's always best to connect the garage with the minimum ~d_i~ with the remaining ~n-1~ garages.

To connect all of them, we need at least ~n-1~ roads (initially there is only one garage, and for each additional one, we need one more road to connect it with the rest). In the optimal solution, we will build exactly ~n-1~ roads because it is never beneficial to connect two garages if there is already a path between them.

Therefore, the total length will be an expression consisting of ~2(n-1)~ terms. Each garage needs to be connected to at least one road, so each of the numbers ~d_1, d_2, \dots, d_n~ must appear at least once in this expression. The smallest possible value of the remaining ~n-2~ terms is precisely the value of the smallest ~d_i~. Notice that this is actually achieved if we connect the garage with the smallest ~d_i~ with the rest.