A functional inequality about log-concave functions Let $f,g$ be  smooth even log-concave functions on $\mathbb{R}^{n}$, i.e.,$f=e^{-F(x)}, g=e^{-G(x)}$ for some even convex functions $F(x),G(x)$. Is it true that:
$$
 \int_{\mathbb{R}^{n}}  \langle \nabla f(x), x\rangle\; g  dx \cdot \int_{\mathbb{R}^{n}}  \langle \nabla g(x), x\rangle\; f  dx \leq \int_{\mathbb{R}^{n}}\langle \nabla f(x), x\rangle\ \langle \nabla g(x), x\rangle\ dx \cdot \int_{\mathbb{R}^{n}}fg\,dx\; \quad (*)  
$$
The answer seems to be positive. 
Motivation: There is an interesting question which asks whether the function $\varphi(t)=\mu(e^{t} K)$ is log-concave on $\mathbb{R}$ where $K$ is a symmetric convex body in $\mathbb{R}^{n}$ and $d\mu$ is  an even log-concave probability measure. Now if you consider the expression $(\ln \varphi(t))''_{t=0}$ and ``do integration by parts'' several times and simplify few expressions you will arrive to an inequality which follows from (*) but definitely is weaker than (*). So I was wondering if there is a simple counterexample to (*).
 A: Not sure whether it is still of interest after three years, but anyway, I suggest the following:
ensure that your assumptions on $F$ and $G$ imply
1. $x\mapsto\langle \nabla F,x\rangle$ (same for $G$) is increasing with respect to some partial order $\le$ on $\mathbb{R}^{d}$
2. $(\mathbb{R}^{d}, \mathcal{B}(\mathbb{R}^{d}), Q, \le)$ with $Q(dx)=\frac{fg}{\langle f,g\rangle}dx$ is an FKG-space, c.f. https://projecteuclid.org/download/pdf_1/euclid.ecp/1465058067.
Hope that helps. Regards
A: edit: I overlooked the word "even" in the question, so my attempted counter-example isn't valid.  Sorry about this.
I don't think the proposed inequality holds.  To see this, consider the following (one-dimensional) situation:

This represents the case where $F$ and $G$ are equal $0$ on an interval each, and grow quickly outside these intervals.
For definiteness, let
$$
F(x) = \begin{cases}
  c(A-x)^2 & \mbox{if $x<A$,} \\
  c(x-B)^2 & \mbox{if $x>B$, and} \\
  0 & \mbox{otherwise,}
\end{cases}
$$
for constants $A$ and $B$, and $G$ be defined similarly, for different
constants $A$ and $B$.  Now consider what happens when we let the constant $c$ go to infinity: In this case, the functions $f$ and $g$ will converge to step functions, while the slope at the edges of the steps goes to $\pm\infty$.
Left hand side:
Since $\int_B^{B+\sqrt{c}} \exp\bigl(-(x-B)^2\bigr) \propto -\sqrt{c}$ the downward slope of $f$ makes a contribution which is asymptotically proportional to $-\sqrt{c}$, and the contribution of the upward slope is negligible in comparison.  By symmetry, the second integral makes the same contribution, and thus the left-hand side is asymptotically proportional to $c$.
Right-hand side: The integral $\int f(x)g(x) \,dx$ converges to a constant and since the derivatives of $f$ and $g$ are never both large at the same time, the first integral on the right-hand side will decay as $c$ increases.
Thus, for this example, as $c\to\infty$, the left-hand side goes to infinity while the right-hand side goes to zero, and thus the proposed bound cannot hold for this example.
