Gebelein's Inequality and convergence of distribution We know that for a bivariate standard normal vector $Z=(Z_1,Z_2)$ it holds that 
\begin{align*}
\operatorname{Cov}(1\{Z_1\leq u),1\{Z_2\leq u))\leq \operatorname{Cov}(Z_1,Z_2).
\end{align*}
This result is known as Gebelein's Inequality. Now we consider two sequences $X_i$ and $Y_i$ $(i\in\mathbb{N})$ of real valued random variables with finite variance and it holds that $X_i$ converges to $Z_1$ in distribution and $Y_i$ to $Z_2$ as $i\rightarrow\infty$. Can we follow that there exists an $l\in\mathbb{N}$ that for all $i>l$
\begin{align*}
\operatorname{Cov}(1\{X_i\leq u),1\{Y_i\leq u))\leq\operatorname{Cov}(X_i,Y_i)
\end{align*}or
\begin{align*}
\operatorname{Cov}(1\{X_i\leq u),1\{Y_i\leq u))\leq \operatorname{Cov}(Z_1,Z_2).
\end{align*}
 A: If you have the stronger hypothesis that the sequence of pairs $(X_n, Y_n)$ converges in law to $(Z_1, Z_2)$ then you'll have that by continuous mapping property** we have that $(1_{X_n \leq u}, 1_{Y_n \leq u})$ converges in law to $(1_{Z_1 \leq u}, 1_{Z_2 \leq u})$. So any expectation-like operator of it will converge also, and thus for any $\varepsilon$ there exist $l$ such that for $n > l$
$$ \operatorname{Cov}(1_{X_n \leq u}, 1_{Y_n \leq u}) \leq \operatorname{Cov}(1_{Z_1 \leq u}, 1_{Z_2 \leq u}) + \varepsilon$$
So for $\varepsilon$ smaller than $\operatorname{Cov}(Z_1, Z_2) - \operatorname{Cov}(1_{Z_1 \leq u}, 1_{Z_2 \leq u}) > 0$ you have your bound.
I don't think you can lift the joint weak convergence, since if you don't have any information about the covariance structure of the sequence.
** : $1_{\cdot \leq u}$ is indeed a discontinuous mapping. But there exists a slightly general version of the continuous mapping theorem that allow measurable functions with discontinuities, such that this set has zero measure under the limit law. See Billingsley's "Convergence of Probability Measures" Th.2.7
