I am reading a [paper][1] where the author uses the following result from matrix analysis but does not explain why it is true nor provide any reference. 

More specifically, let $B \in \mathbb{R}^{p\times d}$ be a semi-orthogonal matrix, i.e $B^\top B = I_d$, and $d < p$. Let $\Sigma$ and $\Delta$ denote two symmetric positive definite matrices such as $\Sigma - \Delta$ is also positive definite. What they claim is that

$$ \log \det \left\vert B^\top \Sigma^{-1} B \right\vert \leq \log \det \left\vert B^\top \Delta^{-1} B \right\vert.$$  

Could someone point me in the direction of explaining why is it true? 

I am thinking of using Poincaré separation theorem, which provides the bounds on the eigenvalues of the matrices on both the left and right-hand sides; however, I did not make any progress with it. 

---

The claim is in the proof of Proposition 3, at the top of page 33. Here $\Sigma = \text{Var}(X)$, and $\Delta = E (\text{Var}(X|y))$, so $\Sigma-\Delta = \text{Var}(E(X \vert y))$  is also positive definite. The $B$ in my question plays the role as $𝐵_0$ in the paper. 

  [1]: http://dx.doi.org/10.1198/jasa.2009.0106