What is $ \pi$ here? It's not $ 3.14...$ I guess, since then the problem is trivial (even for $ n<6$) so what is it then?

Peter wrote: What is $ \pi$ here? It's not $ 3.14...$ I guess, since then the problem is trivial (even for $ n < 6$) so what is it then? Yeah, it is not. Here, $ \pi(x)$ denotes the number of primes $ \leq x$.

For $ n = 7,8,9$ this can easily be checked. So assume $ n\ge 10$. Then we have that $ \sqrt{p_{1}p_{2}\cdots p_{n}}> 3^{n}$. Now, we know that for all positive integers $ m$, there is at least $ 2$ primes in $ ]m,3m]$ (strictly more for $ m>3)$. (I have to use this fact, I see no solution without?) So, there are more than $ 2n$ primes in $ ]1,3]\cup]3,9]\cup\cdots\cup]3^{n-1},3^{n}] = ]1,3^{n}]$, thus $ \pi\left(\sqrt{p_{1}p_{2}\cdots p_{n}}\right)\ge\pi (3^{n})> 2n$.

We can get by using the fact that $ p_{n+1}<2p_n$. It follows that $ p_{2n+2}<4p_{2n}$. For $ n\ge 6$, since $ p_{n+1}\ge 17$, it follows that $ \sqrt{p_{n+1}}>4>\frac{p_{2n+2}}{p_{2n}}$. Thus, if it is true that $ \sqrt{p_1p_2\ldots p_n}>p_{2n}$, it is also true that $ \sqrt{p_1p_2\ldots p_np_{n+1}}>p_{2n+2}$. Since this is indeed true for $ n=6$ ($ \sqrt{30030}>37$), it is true for all $ n\ge 6$ by induction. Luke See my puzzle blog at http://bozzball.blogspot.com

Very nice solution!