Integrated multivariate normal density is greater or equal to product of integrated marginal densities Let $(X, Y)$ be a two-dimensional multivariate normally distributed random vector with means $0$, standard deviations $1$ and correlation $\rho$.
Given the joint density $f_{X, Y}$  of $(X, Y)$ and its marginal densities $f_X, f_Y$, I want to show
$$
\int_u^\infty \int_v^\infty f_{X, Y}(x, y)\ \mathrm{d}x\, \mathrm{d}y \geq 
\bigg( \int_u^\infty  f_{X}(x)\ \mathrm{d}x\ \bigg)
\bigg( \int_v^\infty  f_{Y}(y)\ \mathrm{d}y\ \bigg)
$$
for any $u, v \geq 0$ (possibly also $u, v \in \mathbb{R}$ if that makes no difference).
I already tried to bring everything to the LHS and compute
$$
\int_u^\infty \int_v^\infty
\Big(f_{X, Y}(x, y) - f_{X}(x) f_{Y}(y)\Big)\
\mathrm{d}x\, \mathrm{d}y.
$$
Here, the integrand equals
$$
\frac{1}{2\pi} \bigg( 
  \frac{1}{\sqrt{1 - \rho^2}} \exp\bigg\{- \frac{x^2 + y^2}{2(1 - \rho^2)} + \frac{
\rho xy}{2(1 - \rho^2)}\bigg\} 
-
  \exp\bigg\{- \frac{x^2 + y^2}{2}\bigg\}
\bigg).
$$
I am not entirely sure how to go from here or if this leads somewhere. Any ideas how to proceed in this calculation or maybe use some other properties from the multivariate normal distribution to show the inequality?

Your claim is true if and only if $\rho\geq 0$. Indeed, for all $u, v\in\mathbb R$, note that
$$\mathbb P(X\geq u, Y\geq v) = \mathbb P(X\leq -u, Y\leq -v)\,,$$
as $(X, Y)$ has the same law as $(-X, -Y)$. Now, the RHS above is the CDF of a bivariate normal with correlation $\rho$. We denote it as $F_\rho(u, v)$, so that we can write
$$F_\rho(u, v)=\mathbb P(X\geq u, Y\geq v)\,.$$ Now, if $\rho=0$, $X$ and $Y$ are independent. We thus have that
$$F_{\rho=0}(u, v) = \mathbb P(X\geq u)\mathbb P(Y\geq v)\,.$$
To conclude, note that $\rho\mapsto F_\rho(u, v)$ is a strictly increasing map (see for instance Bivariate normal distribution and correlation). In particular, we have that for $\rho>0$
$$F_{\rho}(u, v)>\mathbb P(X\geq u)\mathbb P(Y\geq v)\,,$$
while for $\rho<0$
$$F_{\rho}(u, v)<\mathbb P(X\geq u)\mathbb P(Y\geq v)\,.$$
