I have no specific reason other than that I know that the candidate is a positive definite function whose derivative is a function of all the states.

I see. Here is a hint. Consider the (autonomous) planar ODE:

\[

\dot x =

\begin{pmatrix}

x_2 \\

h(x_1)

\end{pmatrix}

\]

where $h$ is continuous. This always produces an integrable system with the Hamiltonian given by

\[

H(x) = \frac{1}{2} x_2^2 + \int_0^{x_1} h(s) \ ds

\]

which can just be checked directly via the fundamental theorem of calculus. Now, if we add some friction as a linear function, the ODE becomes

\[

\dot x =

\begin{pmatrix}

x_2 \\

-\gamma x_2 + h(x_1)

\end{pmatrix}

\qquad \gamma > 0

\]

and the system is now dissipative. The proof of this is very easy since the Hamiltonian in the integrable case is precisely the Lyapunov function required. Also note that all equilibria are preserved but in the old case they were all center type and in the new system they are all stable.

Now, a well known example of this setting is the damped pendulum which is the case where $h(x) = -\sin(x_1)$ (after normalizing the mass). If we also take $\gamma = 1$ we obtain

\[

\dot x =

\begin{pmatrix}

x_2 \\

-x_2 - \sin(x_1)

\end{pmatrix}

\]

and the Lyapunov function is given by $V(x) = -\frac{1}{2} x_2^2 + (1 - \cos(x_1))$.

Notice that this vector field is extremely similar to your situation. In particular, the nonlinear part of your system occurs only in the 2nd and 3rd coordinates, each of which is a function of only the 1st coordinate.