Studying again orbital maneuvers, I found the following plotting the cost of a Hohmman transfer between coplanar circular orbits (black line), in terms of $\Delta V$ against the orbit ratio $r_f/r_i$.
It surprises me to found a maximum $\Delta V$ at about $r_f/r_i=15.58$. As it can be seen in the figure, there is a range of $r_f/r_i$ values above the escape velocity cost, $\sqrt{2}$. For example, reaching a geostationary orbit from LEO (assuming $h=300$ km), $r_f/r_i=6.3139$ is more or less the same cost to put a spacecraft in a Moon's alike orbit $r_f/r_i=54.2977$.
Is there some intuitive explanation to this "apparently" counter-intuitive behaviour?.
Plot source: https://www.reddit.com/r/KerbalSpaceProgram/comments/1ajru7/i_was_curious_about_the_delta_v_requirements_for/
NOTE: only Earth gravity is being considered to take the plot
BONUS: the black line function can be defined mathematically as follows; The needed impulses are
$\Delta V_1=\sqrt{\frac{2 \mu}{r_i}-\frac{2 \mu}{r_i+r_f}}-\sqrt{\frac{\mu}{r_i}}$,
$\Delta V_2=\sqrt{\frac{\mu}{r_f}}-\sqrt{\frac{2 \mu}{r_f}-\frac{2 \mu}{r_i+r_f}}-$,
being the total impulse of the Hohmman transfer given by $\Delta V=\Delta V_1 + \Delta V_2$. If one divides the velocity increments by the initial velocity $V_i=\sqrt{\mu/r_i}$ it is obtained
$\frac{\Delta V_1}{V_i}=\sqrt{2-\frac{2r_i}{r_i+r_f}}-1$,
$\frac{\Delta V_2}{V_i}=\sqrt{\frac{r_i}{r_f}}-\sqrt{\frac{2r_i}{r_f}-\frac{2r_i}{r_i+r_f}}$,
defining $\lambda=r_f/r_i$ (note that this is the variable of the x-axis of the plot), and operating
$\frac{\Delta V_1}{V_i}=\sqrt{\frac{2\lambda}{1+\lambda}}-1$,
$\frac{\Delta V_2}{V_i}=\sqrt{\frac{1}{\lambda}}-\sqrt{\frac{2}{\lambda(1+\lambda)}}$,
and the total cost expressed as a function of $\lambda$ is
$\frac{\Delta V}{V_i}=\frac{\Delta V_1}{V_i}+\frac{\Delta V_2}{V_i}=\sqrt{\frac{2\lambda}{1+\lambda}}(1-\frac{1}{\lambda})+\sqrt{\frac{1}{\lambda}}-1$,
which is the analytical expression of the black line in the plot.