On the concavity of a log-det function Let $X$, $P$ be $n\times n$ real positive semi-definite matrices. Further, let $\{A_k\}_{k=1}^N$ be a sequence of real $n\times n$ matrices and $\alpha$ a real positive scalar.

Question. Is 
  $$
\begin{split}
f(X):=\log \det \left(P\left(\sum_{k=1}^N A_k X A_k^\top +X \right) +\alpha I\right)\\ -\log \det \left(P\left(\sum_{k=1}^N A_k X A_k^\top \right) +\alpha I\right)
\end{split}
$$
  a concave function of $X$?


Addendum. In the scalar ($n=1$) case $f(X)$ has the form
$$
f(X) = \log\frac{(P+\gamma)X+\alpha}{\gamma X+\alpha},
$$
where $\gamma:=P\sum_{k=1}^N A_k^2$. It is easy to see that in this case $f(X)$ is a concave function of $X\ge0$.
 A: We may assume $\alpha=1$. That follows is a proof in the particular case when $A_k=(1/\sqrt{N})I$. The generalization will be your business.
We assume that $X>0,P>0$. The case $P$ semi definite can be deduced by continuity; yet, the hypothesis $X>0$ is necessary, because I have to use the tangent space to $S^+$ (the set of $>0$ matrices) which is $S$, the vector space of symmetric matrices.
Let $f(X)=\log(\det(2PX+I))-\log(\det((PX+I))=g(X)-h(X)$.
$Dh_x:H\in S\rightarrow tr(PH(PX+I)^{-1})$,
$D^2h_X(H,H)=-tr(PH(PX+I)^{-1}PH(PX+I)^{-1})=-tr((PH(X+P^{-1})^{-1}P^{-1})^2)$ and
$D^2h_X(H,H)=-tr((H(X+P^{-1})^{-1})^2$.
In the same way,
$D^2g_X(H,H)=-tr((H(X+1/2P^{-1})^{-1})^2)$.
Note that $0\leq X+1/2P^{-1}\leq X+P^{-1}$ implies that $Q=(X+1/2P^{-1})^{-1}\geq R=(X+P^{-1})^{-1}\geq 0$.
Then, it remains to show that $Q=[q_{i,j}]\geq R=[r_{i,j}]\geq 0$ implies that 
for every $H,\;tr((HQ)^2)\geq tr((HR)^2)$. We fix $H$.
We may assume that $H=diag((h_i)_i)$, where $h_i\in\mathbb{R}$ and we put $h=[h_1,\dots,h_n]^T$.
Thus $tr((HQ)^2)=\sum_{i,j}h_i{q_{i,j}}^2h_j=h^T(Q\circ Q)h$, where $\circ$ is the Hadamard's product. In the same way, $tr((HR)^2)=h^T(R\circ R)h$. 
$Q\geq R$ implies $Q\circ Q\geq R\circ R$ and we are done.
