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Carlo Beenakker
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There is a proof by the author of a stronger inequality.
First follow the argument at the top of page 486 of Hypocoercive diffusion operators to arrive at

Then follow the proof in arXiv:math/0609050, page 152, where it is shown that the functional
evolves in time such that


Alternatively, for the case $n=1$, $V=0$ there is a worked-out proof in page 10 and following of Hérau's lecture notes.
Let me summarize the key steps. As a short-hand notation we write $||\partial_x h||^2=\int (\partial h/\partial x)^2\,d\mu$ and $\langle\partial_x h,\partial_v h\rangle=\int (\partial h/\partial x)(\partial h/\partial v)\,d\mu$. We will make use of the identities $$\langle f,\partial_v g\rangle=\langle(-\partial_v+v)f,g\rangle,$$ $$\langle v\partial_x f,f\rangle=0$$ $$[\partial_v v,\partial_x]=1,$$ and the Cauchy-Schwartz inequality $$2|c\langle\partial_v f,\partial_x f\rangle|\leq c^2||\partial_v f||^2+||\partial_x f||^2.$$ The Fokker-Planck equation for $n=1$, $V=0$ reads $$\partial_t h+v\partial_x h+(-\partial_v+v)h=0.$$ This implies the derivatives $$-\frac{1}{2}\frac{d}{dt}||h||^2=||\partial_v h||^2,$$ $$-\frac{1}{2}\frac{d}{dt}||\partial_x h||^2=||\partial_v\partial_x h||^2,$$ $$-\frac{1}{2}\frac{d}{dt}||\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||(-\partial_v +v)\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||\partial_v^2 h||^2+||\partial_v h||^2.$$

So the derivative $dI/dt$ in the OP reduces to $$\frac{dI}{dt}=-2a||\partial_v\partial_x h||^2-2c\biggl(\langle\partial_x h,\partial_v h\rangle+||\partial_v^2 h||^2+||\partial_v h||^2\biggr)$$ $$\qquad\leq -2a||\partial_v\partial_x h||^2-2c||\partial_v^2 h||^2-(2c-c^2)||\partial_v h||^2+||\partial_x h||^2.$$ It remains to bound $||\partial_x h||^2$. In the lecture notes they do this by adding the mixed term $b\langle\partial_x h,\partial_v h\rangle$ to the left-hand-side of the inequality, which then gives a term $-2b||\partial_x h||^2$ on the right-hand-side to dominate.

Carlo Beenakker
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