= Solution
The material surface is $F(x,y,t)=y-\eta(x,t)=0$. Its material derivative vanishes, giving the exact kinematic condition
$$
\boxed{\eta_t+\phi_x\eta_x=\phi_y\quad\text{on }y=\eta.}
$$
For <irrotational flow> $u=\nabla\phi$, the <Euler equations> and $(u\cdot\nabla)u=\nabla(|u|^2/2)$ integrate spatially to $\phi_t+|\nabla\phi|^2/2+gy+p/\rho=C(t)$. Adding a suitable function of time to $\phi$ changes no velocity and makes the right side $p_a/\rho$. Thus
$$
\boxed{\phi_t+\frac12|\nabla\phi|^2+gy+\frac{p-p_a}{\rho}=0.}
$$
Take $p_a$ as the air-side pressure and use the given capillary pressure jump on the fluid side. At first order in wave amplitude, the product $\phi_x\eta_x$ and kinetic-energy term are quadratic, the curvature denominator is one to the required order, and replacing $y=\eta$ by $y=0$ changes first-order quantities only at second order. Consequently
$$
\boxed{\phi_y=\eta_t,\qquad\phi_t+g\eta-\frac T\rho\eta_{xx}=0\quad(y=0).}
$$
For a harmonic wave $\eta=A\cos(kx-\omega t)$, the harmonic potential decaying with depth is
$$
\phi=\frac{A\omega}{k}e^{ky}\sin(kx-\omega t),\qquad y<0.
$$
It satisfies the kinematic condition. Substituting into the dynamic condition gives
$$
\boxed{\omega^2=gk+\frac T\rho k^3.}
$$
The phase speed is $c=\omega/k$ and the <group velocity> of a narrow wave packet is $c_g=d\omega/dk$. For the positive-frequency branch and $\sigma=T/\rho$,
$$
c_g=\frac{g+3\sigma k^2}{2\sqrt{gk+\sigma k^3}},\qquad
\frac{c_g}{c}=\frac{g+3\sigma k^2}{2(g+\sigma k^2)}.
$$
For positive <surface tension> this ratio exceeds one exactly when $\sigma k^2>g$, proving
$$
\boxed{c_g>c\quad\Longleftrightarrow\quad k>\sqrt{\rho g/T}.}
$$
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