[ Inequality problems. Case analysis]

Auto: Cheboraraev $A$.

In a store with three floors, movement between them is possible only by elevator. A study of floor visitation showed that from the start of the working day until the store closes:

1) of the customers entering the elevator on the second floor, half go to the first floor, and half - to the third;
2) among the customers exiting the elevator, less than a third do so on the third floor.

To which floor did customers travel more often from the first floor, to the second or to the third?
Suppose that during the entire day, $x$ customers entered the elevator on the first floor, $-y$ on the second floor, and $-z$ on the third floor. Note that the number of customers exiting the elevator on each floor is equal to the number of customers entering on that same floor.

According to the condition, half of the customers entering on the second floor go down, and half go up. This means that $y / 2$ customers travel from the second floor to the third floor, and the same number travel from the second floor to the first floor. The second condition can be written as: $z < 1 / 3(x + y + z)$. This is equivalent to $2z < x + y$.

From the first floor to the third floor, there were $z - y / 2$ trips, since a total of $z$ people exited the elevator on the third floor, and $y / 2$ of them came from the second floor. Those who went from the first floor to the second floor were the customers entering the elevator on the first floor who did not go to the third floor, so there were $x - (z - y / 2)$ of them. We need to compare these two expressions. However, the inequality $z - y / 2 < x - (z - y / 2)$ is equivalent to the already proven inequality $2z < x + y$. Thus, we have proven that fewer customers traveled from the first floor to the third floor than from the first floor to the second floor on that day.

## Answer

From the first to the second.