TITLE: A conjecture about the prime function $p_n$: $p_m \cdot p_n >p_{m \cdot n}$
QUESTION [10 upvotes]: While testing my system Zet for computational mathematics I find possible relations now and then. The latest is:
Conjecture:

For all $(m,n)\in\mathbb Z_+^2$ except $(3,4),(4,3) \text{ and } (4,4)$ it
  holds that $p_m\cdot p_n > p_{m\cdot n}$

REPLY [4 votes]: The proof of the conjecture for sufficiently large $n,m$ has been given in the other answer. I will here focus on getting control on the "sufficiently large" part thereby giving a way to make a proof of the conjecture for all $n,m$. I will stick to simply presenting the inequalities and the range for which they are satisfied without showing all the hairy calculations so you should try to verify these for yourself.

For $n\geq 6$ Rosser's theorem gives precice inequalities for the $n$'th prime number
$$\log(n) + \log\log(n) - 1 < \frac{p_n}{n} < \log(n) + \log\log(n)$$
From this we get $[\log(n) +\log\log(n)-1][\log(m)+\log\log(m)-1] < \frac{p_np_m}{nm}$ and $\frac{p_{nm}}{nm} < \log(nm) + \log\log(nm)$ so the conjecture follows if we can prove
$$\log(nm) + \log\log(nm) < [\log(n) +\log\log(n)-1][\log(m)+\log\log(m)-1]$$
This inequality can be shown to hold for all $n,m\geq 15$. The cases $n,m<15$ can be done by brute force and shows only three cases where it's violated: $(3,4),(4,3)$ and $(4,4)$. The remaining cases follows if we can prove
$$\log(mn) + \log\log(mn) < \frac{p_m}{m}[\log(n) + \log\log(n)-1]$$
for $m=1,2,3,\ldots,14$ for $n\geq 15$. This inequality can again be shown to hold for $n\geq 33$. A final brute-force check of the cases $n\in[15,33]$ and $m\in[1,14]$ completes the proof of the conjecture.

Note that one does not need to prove the best possible version of the inequalities above (i.e. $15$ for the first inequality and $33$ for the second inequality). We can strick with a worse value at the expence of checking more cases numerically. Replacing $15$ and $33$ by $100$ and $1000$ should make the proof of the inequalities easier and it will still be fast to verify the remaining cases $(n \leq 100, m \leq 1000)$ numerically.