coordination compounds - Splitting of $d$ orbitals when ligands approach central metal ion

In my high school chemistry book, it is written that when ligands approach the central metal ion (transition metal ion) to form dative bonds, the $3d$ orbitals split into two: two which are in higher energy level and the other 3 in lower energy orbital...

One thing that doesn't make intuitive sense to me is that why 3 orbitals move to a lower energy level... if the ligands repel the electrons in the $3d$ orbitals, how can some of the orbitals possibly lose energy and be at a lower enrgy state than they were currently at before...

Answer

The picture you pose is only a part. The full picture is posted below.

This solves your doubt. The splitting does not occur about the original level, i.e. not about the energy level the d-orbital has in complete absence of the ligand's electrostatic field. But, the splitting occurs about a hypothetical Barycentre. This is the energy level of the d-orbitals (still degenerate) in a hypothetical spherical electric field (with spherical symmetry). Due to the symmetry of the hypothetical barycentre field, the d-orbitals remain degenerate but at an energy higher than in the absence of the ligands. Now, the $t_{2g}$ group which is lowered in case of an octahedral field, is still above the original free ion energy level in the complete absence of the ligands.

EDIT
By conservation of energy, we mean that the energy is conserved during splitting. The very process of splitting occurs symmetrically about the barycentre, so that the net amount by which $t_{2g}$ is lowered ($3\times 0.4\Delta_o$) is equal to the amount by which $e_g$ is raised ($2\times 0.6\Delta_o$). But the net energy in absence of field and in presence of field need not be same for energy conservation to hold. Whatever the rise in the net energy (from free ionic to barycentre) will be taken from the kinetic energy (or potential) of the approaching ligands.