Bounding game .Assume for country and WLOG assume that $x_1 \le x_2 \le \dots \le x_n$.We should have $x_n^2|x_1^3+x_2^3+\dots x_{n-1}^3$ and so $x_n^2 \le x_{n-1}^3(n-1)$ and now note that $(x_1x_2 \dots x_n)^2 \ge 4^{n-2} (x_{n-1}x_{n})^2$.Combining this with the inequality we had for $x_n$ it is easy to see $(x_1x_2...x_n)^2 > x_1^3 + x_2^3 +...+x_n^3$ which is a contradiction.

My solution is somewhat different, Assume the contrary like Taha's solution and that $$x_1 \le x_2 \le \dots \le x_n \quad \implies x_n^2|x_1^3+x_2^3+\dots x_{n-1}^3 \quad\text{and} \quad(x_1x_2 \dots x_n)^2 \leq nx_n^3 $$So we get that: $$x_n^2\leq x_1^3+x_2^3+\dots x_{n-1}^3 \quad \text{and}\quad (x_1x_2 \dots x_{n-1})^2 \leq nx_n$$This implies$$\frac{ (x_1x_2 \dots x_{n-1})^4}{n^2}\leq x_1^3+x_2^3+\dots x_{n-1}^3$$Buts it is easy to see that with $n>3$ (the problem is easy with $ n=3$): $$\frac{ (x_1x_2 \dots x_{n-1})^4}{n^2}\ge \frac{2^n(x_1x_2 \dots x_{n-1})^3}{n^2} \ge (x_1x_2 \dots x_{n-1})^3 $$$\implies (x_1x_2 \dots x_{n-1})^3 \leq x_1^3+x_2^3+\dots x_{n-1}^3$ we can see that this is wrong with the fact that $x_1,x_2,...,x_n>1$ and $\forall a,b>1: ab\ge a+b$ $\implies (x_1x_2 \dots x_{n-1})^3 \ge x_1^3+(x_2 \dots x_{n-1})^3 \ge \dots\ge x_1^3+x_2^3+\dots x_{n-1}^3$ And we are done

who is the proposer of this problem? the N2 in shortlist 2019 is an special one of this problem

math-helli wrote: who is the proposer of this problem? the N2 in shortlist 2019 is an special one of this problem Mr.Jamali

Alireza_Amiri wrote: Buts it is easy to see that with $n>3$ (the problem is easy with $ n=3$): $$\frac{ (x_1x_2 \dots x_{n-1})^4}{n^2}\ge \frac{2^n(x_1x_2 \dots x_{n-1})^3}{n^2}$$ Why??

Similar problem IMO shortlist 2019 N2

Solved with ihategeo_1969. Maybe there's an explanation but I found it funny that this problem appeared on the Iran MO the exact same year it was on the shortlist (and not the year after). In the typical size spirit, we just do crazy bounding. WLOG, assume that $x_1 \le x_2 \le \dots \le x_n$. Then, since $x_n^2$ divides the left hand side, it must also divide the right and thus, \[x_n^2 \mid x_1^3 +\dots + x_{n-1}^3\]From this it then follows that, $x_n^2 \le x_1^3 +\dots x_{n-1}^3$. Now, we also note that, \[nx_n^3 \ge x_1^3 +\dots + x_n^3 = (x_1\dots x_{n})^2\]so, $nx_n \ge (x_1\dots x_{n-1})^2$. Combining this with our previous bound we have, \[\left( \frac{(x_1\dots x_{n-1})^2}{n}\right)^2\le x_n^2 \le x_1^3 +\dots x_{n-1}^3\]Further, \[(n-1)x_1^3 \ge x_1 ^3 + x_2^3 + \dots + x_{n-1}^3 \ge \frac{(x_1x_2\dots x_{n-1})^4}{n^2} \ge \frac{2^{4n-8}x_1^4}{n^2} \ge \frac{2^{4n-7}x_1^3}{n^2}\]Thus, $n^2(n-1) \ge 2^{4n-7}$ which is most clearly not true for any positive integer $n \ge 3$ as desired.

We can just directly compare LHS with RHS.