 Solution

For phase $kx-\omega t$, a constant-phase point moves at the phase velocity $\boxed{v_P=\omega /k}$. A narrow-band wave packet has envelope speed given by the local group velocity $\boxed{v_G=d\omega /dk}$ on its dispersion relation.

Since $\omega =k{v}_P(k)$,Use the in-medium wavelength $\lambda =2\pi /k$, so $d\lambda /dk=-\lambda /k$. The velocity relation becomesThe stated increase of phase velocity with wavelength means $d{v}_P/d\lambda >0$. Therefore $\boxed{v_G<v_P}$. The inequality alone does not provide their numerical values or guarantee a positive group speed for an arbitrary dispersion law; it fixes their ordering. For ordinary transparent glass in its normal-dispersion band, both are positive and the group speed is the smaller one.