There is a worked-out proof in page 10 and following of <A HREF="https://hal.archives-ouvertes.fr/hal-01616979/document">Hérau's lecture notes.</A> 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 <A HREF="https://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality">Cauchy-Schwartz</A> + <A HREF="https://en.wikipedia.org/wiki/Young%27s_inequality_for_products#Elementary_case">Young inequality</A>
$$2|c\langle\partial_v f,\partial_x f\rangle|\leq 2c \|\partial_v f \|\, \|\partial_x f \|\leq c^2 \|\partial_v f \|^2+ \|\partial_x f \|^2.$$
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}