= 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 $\omega=kv_P(k)$,
$$
v_G=v_P+k\frac{dv_P}{dk}.
$$
Use the in-medium <wavelength> $\lambda=2\pi/k$, so $d\lambda/dk=-\lambda/k$. The velocity relation becomes
$$
\boxed{v_G=v_P-\lambda\frac{dv_P}{d\lambda}}.
$$
The stated increase of <phase velocity> with <wavelength> means $dv_P/d\lambda>0$. 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.
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