Variance Inequality for Random Vectors I know that if X and Y are random scalar variables, then:
\begin{align*} 
\mathrm{Var}(X+Y) & = \mathrm{Var}(X) + \mathrm{Var}(Y) +  2\mathrm{Corr}(X,Y)\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)} \\
                  & \le \left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)^2
\end{align*}
I would like to know if the same inequality holds for random vectors, i.e.: $X , Y \in \mathcal{R^{n}}$
\begin{align*} 
\mathrm{Var}(X+Y) & = \mathrm{Var}(X) + \mathrm{Var}(Y) +  \mathrm{Cov}(X,Y)+ \mathrm{Cov}(X,Y)' \\
                  & \le \left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)\left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)'
\end{align*}
Where square root of a variance (positive definite) matrix is its Cholesky decomposition. And the inequality means that the difference is semidefinite positive. 
As an extra hypothesis, I have that both $\mathrm{Var}(X)$ and $\mathrm{Var}(Y)$ are  diagonal matrices. 
I have been looking for Cauchy Schwartz inequalities for variance matrices, and I found this: 
\begin{equation}
\mathrm{Var}(Y) \ge \mathrm{Cov}(Y,X) \mathrm{Var}(X)^{-1} \mathrm{Cov}(X,Y)
\end{equation} But I don't know how to give any use to it. Thanks a lot!
 A: This doesn't hold. Take this example: Let $X, \, Y \in \mathcal{R}^2$ and the variance matrix of $\left( \begin{matrix} 
   X \\
   Y
 \end{matrix} \right)$ is
$D =\left( \begin{matrix} 
   250  & 150 &  155 &  115 \\
   150  & 100 &   95 &   80 \\
   155  &  95 &  133 &   87 \\
   115  &  80 &  87  &  70 \\ \end{matrix} \right) $. 
Which is positive definite with eigenvalues : 0.0029,  14.3177 29.2519 and 509.4275.
Now, $\mathcal{Var}(X+Y) = \left(\begin{matrix} 693 & 447 \\ 447 &330\end{matrix}\right)$ and 
$\sqrt{\mathrm{Var}(X)}=\text{Chol}\left[ \left(\begin{matrix} 250  & 150  \\ 150  & 100 \end{matrix}\right)\right] = \left(\begin{matrix} 15.8114  &  0   \\  9.4868   &  3.1623 \end{matrix}\right)$
$\sqrt{\mathrm{Var}(Y)}=\text{Chol}\left[ \left(\begin{matrix} 133 &   87 \\ 150  & 100 \end{matrix}\right)\right] = \left(\begin{matrix} 11.5326      &  0   \\  7.5439 &   3.6180 \end{matrix}\right)$
Sadly, $\left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)\left(\sqrt{\mathrm{Var}(X)} + \sqrt{\mathrm{Var}(Y)}\right)' - \mathrm{Var}(X+Y)  = \left(\begin{matrix} 54.6917  & 18.6863  \\ 18.6863  & 6.0171  \end{matrix}\right) $ 
has -0.3292 as eigenvalue, so it's not positive definite.  
