7. Suppose ten people each holding a bucket come to a water tap to fill their buckets. Let the time required to fill the bucket of the $i$-th person $(i=1,2, \cdots, 10)$ be $T_{i}$ minutes. Assume these $T_{i}$ are all different. 

When only one water tap is available, how should the order of these ten people be arranged to minimize their total waiting time (including the time each person spends filling their own bucket)? What is this minimum total time? (You must prove your argument).
7. If the buckets are filled in the order of $i_{1}, i_{2}, \cdots, i_{10}$, then the total time required is
$$
f(t)=T_{i_{1}}+\left(T_{i_{1}}+T_{i_{2}}\right)+\cdots+\left(T_{i_{1}}+T_{i_{2}}+\cdots+T_{i_{10}}\right)=10 T_{i_{1}}+9 T_{i_{2}}+\cdots+2 T_{i_{9}}+T_{i_{10}} \text {. }
$$

Assuming $T_{1}<T_{2}<\cdots<T_{10}$, then according to the rearrangement inequality, we have $f(t) \geqslant 10 T_{1}+9 T_{2}+\cdots+T_{10}$, the minimum time spent is $10 T_{1}+9 T_{2}+\cdots+T_{10}$, and the order of the ten people should be arranged according to the principle of those who take less time to fill the bucket going first.