Yes, we may achieve even $\nu(n)\leqslant 2\min(n,\pi(n))$. Indeed, define $\rho(n)=\min(n,\pi(n))$ and let $t_1,t_2,\dots$ be an enumeration of the positive integers such that $\rho(t_1)\leqslant \rho (t_2)\leqslant \rho(t_3)\leqslant \dots$. Then $\rho(t_k)\geqslant k/2$, otherwise we may find three different numbers with the same values of $\rho$. Therefore defining $\nu(t_k)=k$ we get $\nu(t_k)=k\leqslant 2\rho(t_k)$ as desired.
This results in $$ \frac{1}{n} + \frac{1}{\pi(n)} \leq \frac{2}{\rho(n)} \leq \frac{4}{\nu(n)}, $$ as was to be shown.
