From the operator monotonicity of the logarithm it follows for positive semidefinite $B, X$
$\log(X+B) \geq \log(X)$.
Does from this we can conclude that for any positive semidefinite $A$
$\text{Tr}(A\log(X+B)) \geq \text{Tr}(A\log(X))$?
One can assume that $A, B, X$ are Hermitian if it helps.
Best Answer
Yes. In general, if $S\ge T$ and $A\ge0$, then $A^{1/2}SA^{1/2}\ge A^{1/2}TA^{1/2}$ and hence $$ \operatorname{tr}(AS)=\operatorname{tr}(A^{1/2}SA^{1/2})\ge\operatorname{tr}(A^{1/2}TA^{1/2})=\operatorname{tr}(AT). $$