There is a worked-out proof in page 10 and following of Hérau's lecture notes. The detailed steps are for $n=1$, $V=0$, but I assume once that is understood, the more general case would follow smoothly.
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 $$\langle f,\partial_v g\rangle=\langle(-\partial_v+v)f,g\rangle,$$ $$\langle v\partial_x f,f\rangle=0$$ $$[\partial_v ,v\partial_x]=\partial_x,$$ and the Cauchy-Schwartz + Young inequality $$2|c\langle\partial_v f,\partial_x f\rangle|\leq 2c \|\partial_v f \|\, \|\partial_x f \|\leq c^2 \|\partial_v f \|^2+ \|\partial_x f \|^2.$$ The Fokker-Planck equation for $n=1$, $V=0$ reads $$\partial_t h+Lh=0,\;\;L=v\partial_x + (-\partial_v+v)\partial_v .$$ The adjoint of $L$ is $$L^\ast=-v\partial_x + (-\partial_v+v)\partial_v.$$
The resulting derivatives are $$-\frac{1}{2}\frac{d}{dt} \|h \|^2= \|\partial_v h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt} \|\partial_x h \|^2= \|\partial_v\partial_x h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt} \|\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|(-\partial_v +v)\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2.$$
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 $-b \|\partial_x h \|^2$ on the right-hand-side to dominate. Let me work that out, using the derivative$^{\ast}$ $$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2-2\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-\langle\partial_x h,\partial_v h\rangle,$$ hence $$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial_x h \|^2+b\langle\partial_x h,\partial_v h\rangle+c \|\partial_v h \|^2\biggr)=$$ $$\qquad=-2(c+1) \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c\|\partial_v\partial_v h\|^2-2b\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$ The first four terms on the right-hand-side have a fixed sign, the last two terms can be bounded by the Cauchy-Schwartz + Young inequality.
$^\ast$ The result for $(d/dt)\langle\partial_x h,\partial_v h\rangle$ given on page 10 of the cited lecture notes is mistaken. Here is a derivation: \begin{align} \frac{d}{dt}\langle\partial_x h,\partial_v h\rangle&=-\langle\partial_x Lh,\partial_v h\rangle-\langle\partial_x h,\partial_v Lh\rangle\\ &=-\langle\partial_x h,(L^\ast\partial_v +\partial_v L)h\rangle,\\ L^\ast\partial_v +\partial_v L&=\bigl(-v\partial_x+(-\partial_v+v)\partial_v\bigr)\partial_v+\partial_v\bigl(v\partial_x+(-\partial_v+v)\partial_v\bigr)\\ &=\partial_x+\partial_v+2(-\partial_v+v)\partial_v\partial_v,\\ \Rightarrow\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle&=-\|\partial_x h\|^2-\langle\partial_x h,\partial_v h\rangle-2\langle\partial_x h,(-\partial_v+v)\partial_v\partial_v h\rangle\\ &=-\|\partial_x h\|^2-\langle\partial_x h,\partial_v h\rangle-2\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle. \end{align}