TITLE: Cauchy-Schwarz matrix inequality for random vectors
QUESTION [6 upvotes]: If $X$ and $Y$ are random scalars, then Cauchy-Schwarz says that
$$| \mathrm{Cov}(X,Y) | \le \mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2}.$$
If $X$ and $Y$ are random vectors, is there a way to bound the covariance matrix $\mathrm{Cov}(X,Y)$ in terms of the matrices $\mathrm{Var}(X)$ and $\mathrm{Var}(Y)$? 
In particular, is it true that
$$\mathrm{Cov}(X,Y) \le \mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2},$$
where the square roots are Cholesky decompositions, and the inequality is read as meaning that the right hand side minus the left hand side is positive semidefinite?

REPLY [6 votes]: There is a generalization of Cauchy Schwarz inequality from Tripathi [1] that says that:
\begin{equation}
\mathrm{Var}(Y) \ge \mathrm{Cov}(Y,X)\mathrm{Var}(X)^{-1}\mathrm{Cov}(X,Y)
\end{equation}
in the sense that the diference is semidefinite positive.
He actually says that a student asked about it and couldn't find any other reference (1998!).
[1]: G. Tripathi, ”A matrix extension of the Cauchy-Schwarz inequality”, Economics Letters, vol. 63, nr 1, s. 1–3, apr. 1999, doi: 10.1016/S0165-1765(99)00014-2.