Matrix Log inequality with trace
Solution 1:
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 $$ \operatorname{tr}(AS)=\operatorname{tr}(A^{1/2}SA^{1/2})\ge\operatorname{tr}(A^{1/2}TA^{1/2})=\operatorname{tr}(AT). $$