Solution:
The product of the numbers is $1$, so they cannot all be greater than $1$ or all less than $1$. If all equalled $1$, then the sum would not be greater than the sum of the inverses. So we must have either one or two greater than $1$. Thus it is sufficient to show that we cannot have two of the numbers greater than $1$.

Suppose that $a$, $b > 1$. Then since $a + b + c > 1/a + 1/b + 1/c$, we have $a + b + 1/(ab) > 1/a + 1/b + ab$, and hence $(1 - 1/a)(1 - 1/b) > (a - 1)(b - 1)$. Dividing by $(a - 1)(b - 1)$ gives $ab < 1$. Contradiction.