I suspect that the following matrix inequality is well known, but I can't find a reference or proof:


Given $n \times n$ symmetric matrices $A,B$ such that $I\_n \leq A,B$, is the following true?
$${Tr}\big[\big( \log A^{\frac{1}{2}} B A^{\frac{1}{2}}\big)^p\big] \geq {Tr}\big[\big( \log B \big)^p\big] + {Tr}\big[\big( \log A\big)^p\big].$$


For clarity, for a continuous function $f$ and symmetric matrix $C$, $f(C)$ is simply equal to $U {diag}(f(\lambda\_1),...,f(\lambda\_n))U^t$, where $U$ is a unitary matrix diagonalizing $C$, i.e., $C = U diag(\lambda\_1,...,\lambda\_n)U^t$.


UPDATE: after the discussion with amakelov below, I edited the statement to be more aesthetic. 



Well known matrix inequality?
The desired inequality follows from an majorisation argument : 


According to Bhatia Matrix Analysis Corollary III.4.6 we have :


$log \lambda(AB) \succ log \lambda^\downarrow(A) + log \lambda^\uparrow(B)$. 


Using Corollary II.3.4 from the same book, we conclude that  

$(log \lambda(AB))^p \succ\_w (log \lambda^\downarrow(A) + log \lambda^\uparrow(B))^p
\succ\_w (log \lambda^\downarrow(A))^p + (log \lambda^\uparrow(B))^p$ ,
since $x \mapsto x^p$ is convex and monoton increasing for $x \ge 0$ .


From this it follows the trace inequality.


