As a participant of Polish MO who saw the solution to this problem, I can honestly say, this is one of the hardest and most marvelous problems I've ever come across...

And no one solved it during the contest, even the current IMO participants, so yes, this problem takes patience .

If $m$ is 1, the result is obvious because $n$ must also be 1. Suppose $m\geq 2$. Here if $a_n > n^2$, the inequality can be directly proved. Otherwise we have $n$ positive integers below $n^2$ whose distinct prime factors are at most $m$ in total. From this one can show that $m \gg log(n)$, which again implies the inequality using $a_i \geq i+n-1$. Unfortunately I can't think of an elementary way to prove the crucial claim regarding the size of $m$. The theory of smooth numbers does tell us that in fact $m$ has to be as large as $log^2(n)$.

Definitely, one of the most beautiful problems I have ever seen. I was so impressed when I saw the solution after the contest!

Care posting it here?

Maybe like this?

This problem was proposed by Burii The idea is to write $a_i = p^{k_i} \cdot b_i$, where $p \nmid b_i$ for a prime divisor $p$ of $a_1a_2\ldots a_n$. Then, due to $a_1<a_2<\ldots<a_n<2a_1$ we get that $b_i$ are pairwise distinct. Thus $$b_1b_2\ldots b_n \ge n!$$ Multiplying such inequalities for each $p$ we get $(a_1a_2\ldots a_n)^{m-1} \ge (n!)^{m}$

Beautiful!

Awesome !!

timon92 wrote: This problem was proposed by Burii The idea is to write $a_i = p^{k_i} \cdot b_i$, where $p \nmid b_i$ for a prime divisor $p$ of $a_1a_2\ldots a_n$. Then, due to $a_1<a_2<\ldots<a_n<2a_1$ we get that $b_i$ are pairwise distinct. Thus $$b_1b_2\ldots b_n \ge n!$$ Multiplying such inequalities for each $p$ we get $(a_1a_2\ldots a_n)^{m-1} \ge (n!)^{m}$

Same idea as Imo shortlist 1985, identical

angiland wrote: If $m$ is 1, the result is obvious because $n$ must also be 1. Suppose $m\geq 2$. Here if $a_n > n^2$, the inequality can be directly proved. Otherwise we have $n$ positive integers below $n^2$ whose distinct prime factors are at most $m$ in total. From this one can show that $m \gg log(n)$, which again implies the inequality using $a_i \geq i+n-1$. Unfortunately I can't think of an elementary way to prove the crucial claim regarding the size of $m$. The theory of smooth numbers does tell us that in fact $m$ has to be as large as $log^2(n)$. What's the difference between $>>$ and $>?$

I don't understand is it the product of ${a}_{i}$ or what ?

Just use the result here: https://artofproblemsolving.com/community/c6h364207p2000897