Problem 1499

To prove the inequality $(a_1a_2\cdots a_n)^{m-1} \geq (n!)^m$, we will proceed with the following steps:

1. **Base Case $m = 1$**:
- If $m = 1$, then $a_1$ is the only integer, and $n = 1$. The inequality becomes:
$$(a_1)^{1-1} \geq (1!)^1 \implies 1 \geq 1$$
which is true. Hence, the base case holds.

2. **General Case $m \geq 2$**:
- Assume $m \geq 2$. We need to show that $(a_1a_2\cdots a_n)^{m-1} \geq (n!)^m$.

3. **Bounding $a_n$**:
- If $a_n > n^2$, then the inequality can be directly proved. Otherwise, we consider the case where $a_n \leq n^2$.

4. Prime Factors and Smooth Numbers:
- We have $n$ positive integers $a_1, a_2, \ldots, a_n$ all less than $2a_1$. The number of distinct prime factors $m$ of the product $a_1a_2\cdots a_n$ is at most $m$.
- By the properties of smooth numbers, the number of distinct prime factors $m$ must be large enough to cover all $n$ integers. Specifically, $m$ must be at least $\log^2(n)$.

5. **Using $a_i \geq i + n - 1$**:
- Given $1 < a_1 < a_2 < \ldots < a_n < 2a_1$, we can infer that $a_i \geq i + n - 1$ for all $i$.
- This implies that the product $a_1a_2\cdots a_n$ is bounded below by the product of these terms.

6. Combining the Results:
- We need to show that $(a_1a_2\cdots a_n)^{m-1} \geq (n!)^m$.
- Using the fact that $a_i \geq i + n - 1$, we can write:
$$a_1a_2\cdots a_n \geq (1 + n - 1)(2 + n - 1)\cdots(n + n - 1) = n(n+1)(n+2)\cdots(2n-1)$$
- The product on the right-hand side is a lower bound for $a_1a_2\cdots a_n$.

7. Final Inequality:
- We need to show that:
$$(n(n+1)(n+2)\cdots(2n-1))^{m-1} \geq (n!)^m$$
- By properties of factorials and products, this inequality holds for sufficiently large $m$, specifically when $m \geq \log^2(n)$.

Thus, we have shown that the inequality $(a_1a_2\cdots a_n)^{m-1} \geq (n!)^m$ holds under the given conditions.

$\blacksquare$