Suppose we have a a set of matrices in the complex field of the form $a_iv_iv_i^H$ for $i=\{1,\dots,n\}$ where $a_i$ are positive real scalars and $v_i$ are complex valued finite dimensional matrices all of same dimension. $I$ is the identity matrix. 

Now we want to maximize the following determinant.
$$\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right).$$

I think the matrix which should go on the numerator is the one with the largest product of $\prod_k(1+\lambda_{ik})$ where $\lambda_{ik}$ are the eigenvalues of $a_iv_iv_i^H$. Or in the case of PD which is equal to largest determinant $\det(a_iv_iv^H_i)$. Is this claim right?

So I proceeded:
$$\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right)=\det \left( I+\sum_{j} a_jv_jv_j^H \right) \det \left( \left( I+\sum_{j\neq i} a_jv_jv_j^H \right)^{-1}\right).$$
The matrices $a_iv_iv^H_i$ are PSD and have non negative eigenvalue. Could someone please help to complete the proof.

P.S.: I believe this question is related to http://mathoverflow.net/questions/65424/determinant-of-sum-of-positive-definite-matrices.