I am looking for the proof of the following concave like property of the logarithm operator.
Let $\bf{0} \preceq {\bf A} \preceq {\bf I}$. Then, for any positive definite matrices $ {\bf \Sigma}_1, {\bf \Sigma}_2$ \begin{align} \log \left( ({\bf I} - {\bf A} )^{\frac{1}{2}} {\bf \Sigma}_1 ({\bf I} - {\bf A} )^{\frac{1}{2}} + {\bf A} ^{\frac{1}{2}} {\bf \Sigma}_2 {\bf A}^{\frac{1}{2}} \right) \succeq ({\bf I} - {\bf A} )^{\frac{1}{2}} \log ( {\bf \Sigma}_1 ) ({\bf I} - {\bf A} )^{\frac{1}{2}} + {\bf A} ^{\frac{1}{2}} \log ( {\bf \Sigma}_2) {\bf A} ^{\frac{1}{2}}. \end{align}
Note that this is true for $n=1$ since
\begin{align} \log( (1-\alpha) \sigma_1+\alpha \sigma_2) \ge (1-\alpha) \log(\sigma_1)+\alpha \log(\sigma_2) \end{align}
A reference would be greatly appreciated.
The stated inequality is basically a special case of Hansen and Pedersen's characterisation of operator convex functions. See section 2.5 of
or
It is well known that $\log:(0,\infty)\to\mathbb R$ is operator concave (this is also proved in example 13(iii) of Chansangiam's survey paper). It follows that for every $\epsilon>0$, the function $f_\epsilon(t):[\epsilon,\infty)\to\mathbb R$ defined by $t\mapsto-\log(t+\epsilon)$ is operator convex. Now, by Hansen-Pedersen characterisation (equivalence of conditions (i) and (v) in theorem 2.5.2 of Hiai or theorem 9 or Chansangiam), the stated inequality in your question holds when $\log$ is replaced by $-f_\epsilon$. Let $\epsilon\to0$, the conclusion follows.