= Solution
Move the three coins as follows, with $c_1$ smallest:
|| Step
|| Coin
|| Source
|| Destination
| 1
| $c_1$
| $s_1$
| $s_3$
| 2
| $c_2$
| $s_1$
| $s_4$
| 3
| $c_3$
| $s_1$
| $s_2$
| 4
| $c_2$
| $s_4$
| $s_2$
| 5
| $c_1$
| $s_3$
| $s_2$
Every moved coin is exposed, and the last two moves put a smaller coin on a larger one. Thus
$$
\boxed{5<2^3-1=7\text{ moves}.}
$$
Five is also optimal here: before the first move of $c_3$, both smaller coins must leave $s_1$. Neither can make its only move directly to the final destination and remain there, because it would obstruct the eventual arrival of $c_3$. Each smaller coin therefore needs at least two moves. Those four small-coin moves and at least one move of $c_3$ give the matching lower bound.
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