We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.
If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.