Iranian Mathematical Olympiad

In every moment consider the number of persons that are on the escalator at that time. Now consider the intervals such that in every time of such intervals the number of persons on the escalator is equal to a fixed integer. Suppose that we have $k$ intervals $I_1, I_2, \dots, I_k$ and for $1 \le i \le k$, $a_i$ and $l_i$ denotes the length and number of persons on the escalator in every moment of $I_i$. We claim that $\sum_{i=1}^{k} a_i^{1-\alpha} t_i = nl$.

Since every person have moved a distance equal $l$ so the sum of travelled distance of people is $nl$. On the other hand travelled distance of a person who is on the escalator on the interval $I_i$ equals $a_i^{-\alpha} t_i$. So the sum of travelled distance in the interval $I_i$ equals $a_i \cdot a_i^{-\alpha} t_i = a_i^{1-\alpha} t_i$, hence the total travelled distance is $\sum_{i=1}^{k} a_i^{1-\alpha} t_i$.

So the claim is proved.

Now consider the following cases.

Case 1. $\alpha \ge 1$. If $a_i \ge 1$ since $\alpha \ge 1$ we have $a_i^{\alpha-1} \ge 1$, hence $a_i^{1-\alpha} \le 1$. If $a_i = 0$ also $a_i^{1-\alpha} = 0 \le 1$. So $nl = \sum_{i=1}^k a_i^{1-\alpha} t_i \le \sum_{i=1}^k t_i$. So the required time is at least $nl$. If each person goes on the escalator when the previous one reached the top of the escalator the required time equals $nl$. Hence in this case the required time is at least $nl$.

Case 2. $\alpha < 1$. Since $a_i < n$ and $1 - \alpha > 0$ we have $a_i^{1-\alpha} < n^{1-\alpha}$ so
$$nl = \sum_{i=1}^{k} a_{i}^{1-\alpha} t_{i} \leq \sum_{i=1}^{k} n^{1-\alpha} t_{i} = n^{1-\alpha} \sum_{i=1}^{k} t_{i} \Rightarrow n^{\alpha} l \leq \sum_{i=1}^{k} t_{i}$$
So the required time is at least $n^\alpha l$. If all $n$ people go on the escalator together the velocity equals $n^{-\alpha}$ and so the required time equals $\frac{l}{n^{-\alpha}} = n^\alpha l$. Hence in this case the required time is at least $n^\alpha l$. $\square$