The classical FKG inequality states that for two non-decreasing function $f,g: R \to R$ we have that \begin{align} E[ f(X) g(X) ] \ge E[f(X) ] E[g(X)] \end{align}
I am interested in understanding what is the generalization of this to the following multivariate setting: Consider two functions $f,g: R^n \to R^n$. We want to show the following inequality: \begin{align} E[ \langle g(X), f(X) \rangle ] \ge \langle E[ g(X)], E[f(X)] \rangle ] \end{align}
The issue is that I am not sure what is proper assumption that we can make about $f$ and $g$ since defining increasing function in the multivariate setting is not straight forward.
Perhaps an easy case to consider is when one of the function is a linear \begin{align} E[ \langle g(X), A X \rangle ] \ge \langle E[ g(X)], E[ A X] \rangle \end{align} where $A$ is symmetric positive definite matrix (i.e., so that $f(x) = Ax$ is 'increasing').
The natural definition of increasing function $f: R^n\rightarrow R^n$ is $$\langle x-y, f(x)-f(y)\rangle \geq 0$$for all $x,y$ which is the case when $f$ is the gradient of a convex function since in this case $$\langle x-y, f(x)-f(y)\rangle =\int_0^1\langle x-y, A(t)(x-y)\rangle dt $$ where $A(t)=f'(y+t(x-y)) $ is semipositive definite.
However, even is this favorable case where $f$ and $g$ are the gradients of convex functions, and denoting $B(s)=g'(y+s(x-y)) $ I have doubt that the Tchebychef trick, which was to use
$$\langle g(x)-g(y), f(x)-f(y)\rangle \geq 0$$ works here since
$$\langle g(x)-g(y), f(x)-f(y)\rangle $$$$= \int_0^1\int_0^1\langle B(s)(x-y),A(t)(x-y)\rangle dsdt$$ does not seem positive (recall that $AB+BA$ is not necessarily semi positive definite when $A$ and $B$ are, unless I am mistaken).