= Solution
Put $\tau=T/\rho>0$ and take the positive-frequency branch for $k>0$. The <phase velocity> and <group velocity> are
$$
\boxed{c_p=\frac\omega k=\sqrt{\frac gk+\tau k},\qquad c_g=\frac{d\omega}{dk}=\frac{g+3\tau k^2}{2\sqrt{gk+\tau k^3}}.}
$$
Since the square root is increasing, minimizing $c_p$ is equivalent to minimizing $g/k+\tau k$. Its derivative is $-g/k^2+\tau$, which changes sign from negative to positive at
$$
\boxed{k_c=\sqrt{\frac g\tau}=\sqrt{\frac{\rho g}T}.}
$$
The phase speed diverges at both positive endpoints $k\downarrow0$ and $k\to\infty$, so this stationary point is its global minimum. At $k_c$, $\tau k_c^2=g$, giving
$$
\boxed{c_g(k_c)=c_p(k_c)=\left(\frac{4gT}{\rho}\right)^{1/4}.}
$$
The relative speed is especially transparent:
$$
c_g-c_p=\frac{\tau k^2-g}{2\omega}.
$$
A narrow <wave packet> moves with $c_g$, while its crests move with $c_p$. For $k<k_c$, $c_p>c_g$: \b[crests move forward through the packet], from its rear towards its front. For $k>k_c$, $c_p<c_g$: \b[crests drift backwards relative to the packet], although both absolute velocities are positive. At $k_c$ the leading-order relative drift vanishes. This is the <minimum phase speed of a gravity-capillary wave>.
\Image[/past-exam-of-the-options-i-course-of-the-university-of-oxford/2013-capillary-speeds.png]
{title=Phase and group speeds cross at the minimum phase speed; crest drift relative to the packet reverses there}
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