how do you prove this :
$$ \mu^T \Sigma \mu \cdot \textbf{1}^T \Sigma \textbf{1} \geq (\textbf{1}^T \Sigma \mu)^2 $$
I thought about using C.S. inequality, which yields to :
(writing the matrix $\Sigma$ as $(\sigma_{ij})$ : $$(\sum \sum \sigma_{ij} u_j )^2 \leq (\sum \sum \sqrt {\sigma_{ij} }^2) (\sum \sum \sqrt {\sigma_{ij} }^2 u_j^2 ) $$
i don't know what to do next. I'm afraid i'm missing the cross product of $$ u_i u_j$$
I'm sure it is simple and I'm missing a trivial point. Can you please help me ?
Oh and if you can detail a little bit about the equality case. When does it happen here ?
Your statement as presented is not true for an arbitrary matrix $\Sigma$. For instance, if we take $$ \Sigma = \pmatrix{1&0\\0&-1}, \qquad \mu = \pmatrix{1\\0} $$ then your inequality becomes $0 \geq 1$, which is clearly false.
Presumably, you are meant to assume that $\Sigma$ is a covariance matrix, which is to say that $\Sigma$ is (symmetric and) positive semidefinite. With that assumption, one approach is as follows. To begin, note that there exists a matrix $M$ such that $\Sigma = M^TM$. It follows that $$ \mathbf 1^T \Sigma \mu = \mathbf 1^T M^TM\mu = (M \mathbf 1)^T (M\mu). $$ It now suffices to apply the Cauchy-Schwarz inequality to the vectors $x = M \mathbf 1$, $y = M \mu$.