$\newcommand\R{\mathbb R}$Such pairs of functions may be called comonotone -- cf. comonotone approximation, which, for $n=1$, is an approximation of a piecewise monotonic function by a polynomial with the same monotonicity.
If functions $f$ and $g$ are comonotone in this sense and are in $L^2(\mu)$ for some probability measure $\mu$ over (say) $\mathbb R^n$, then
$$\int fg\,d\mu\ge\int f\,d\mu\,\int g\,d\mu. \tag{1}\label{1}$$
This is Chebyshev's integral inequality -- cf. e.g. this and this.
For completeness, here is a proof of \eqref{1}:
We have $(f(a)-f(b))(g(a)-g(b))\ge0$ for all $a$ and $b$ in $\R^n$, and hence
$$0\le\iint\mu(da)\mu(db)(f(a)-f(b))(g(a)-g(b)) \\
=2\int fg\,d\mu-2\int f\,d\mu\,\int g\,d\mu. \quad\Box$$
More on such comonotonicity: Suppose again that $f$ and $g$ are comonotone, and also suppose that $f$ and $g$ are Borel measurable. Let $Z$ be any random vector in $\R^n$. Let $X:=f(Z)$ and $Y:=g(Z)$. Then it is easy to check that, for any real $x$ and $y$, one of the events $\{X\le x\}$ and $\{Y\le y\}$ is contained in the other one. So, for all real $x$ and $y$
$$P(X\le x,Y\le y)=\min(P(X\le x),P(Y\le y));$$
that is, the random vector $(X,Y)$ is comonotone in this sense.