It's well-known that $$ \liminf_n\frac{\varphi(n)\log\log n}{n}=e^{-\gamma} $$ and there exists an effective version $$ \varphi(n)>\frac {n}{e^\gamma\log\log n+\frac{3}{\log\log n}} $$ valid for $n\ge3.$ Of course the RHS is increasing and so has an inverse, but I would like to know if there is an explicit formula (giving a tight bound) with $$ \varphi(f(n))>n. $$
Is this too much to ask?
Assume we're going to find a bound of the form $n<f(n) < n\log n$ (at least for $n$ large enough) so $$ \begin{align} \log n < \log f & < \log n+\log\log n\\ \log\log n < \log \log f & < \log(\log n + \log\log n)\\ \end{align} $$ Then from your bound $\phi(f)<n$ requires that $$ f < n \left(e^\gamma\log\log f+3/\log\log f\right) \\ f < n \log\log f (e^\gamma+3/(\log\log f)^2) \\ f < n\log(\log n + \log\log n)\left(e^\gamma+3/(\log\log n)^2\right) $$ So $$ N>n\log(\log n + \log\log n)\left(e^\gamma+3/(\log\log n)^2\right) \Rightarrow \phi(N)>n $$ I guess there may be a slightly better version, since the bound for $n=10^7$ gives $6.38\times 10^7$, whereas the last $N$ with $\phi(N)\le 10^7$ seems to be $58198540$.