[proofplan] Use Hamilton's equations in canonical action-angle coordinates. Since the Hamiltonian is independent of the angle variables, the action equations have zero right-hand side, and the angle equations are given by the gradient of $H$ with respect to the action variables. [/proofplan]

[guided] The hypothesis already puts the system in the desired normal coordinates. The proof is therefore a direct substitution into Hamilton's canonical equations: no existence theorem for action-angle coordinates is being proved here. [/guided]

[step: Apply Hamilton's equations] In canonical action-angle coordinates $(I,\theta)$, Hamilton's equations have the form \begin{align*} \dot I_i=-\frac{\partial H}{\partial \theta_i}, \qquad \dot\theta_i=\frac{\partial H}{\partial I_i} \end{align*} for each $i=1,\dots,n$. [/step]

[step: Use the action-only hypothesis] By hypothesis, $H=H(I)$ has no dependence on the angle variables $\theta_i$. Hence \begin{align*} \frac{\partial H}{\partial \theta_i}=0 \end{align*} for every $i$, so Hamilton's first equation gives \begin{align*} \dot I_i=0 \end{align*} for every $i$. [/step]

[step: Identify the angle velocity] Hamilton's second equation gives \begin{align*} \dot\theta_i=\frac{\partial H}{\partial I_i} \end{align*} for every $i$. In vector notation this is \begin{align*} \dot\theta=\nabla_I H(I). \end{align*} Thus the action variables are constant and the angle variables evolve linearly with frequency vector $\nabla_IH(I)$ on each invariant torus. [/step]