Prove that for positive definite matrices $A$ and $B$ where $A – B$ is also positive definite, show
$$2Tr((A-B)^{1/2}) + Tr(A^{-1/2}B) \leq 2Tr(A^{1/2})$$
My attempt so far:
We know that $A – B$ positive definite $\implies Tr(A – B) \geq 0 \implies Tr((A-B)^{1/2}(A-B)^{1/2}) \geq 0$
I'm not sure whether I'm on the right track, or where to go from here. Any direction or solution would be appreciated.
Best Answer
Let $Y=B^{1/2},\,Z=(A-B)^{1/2}$ and $X=A^{1/2}=(Y^2+Z^2)^{1/2}$. Then $X,Y,Z\succ0$ and \begin{aligned} &\operatorname{tr}\left(2A^{1/2}-2(A-B)^{1/2}-A^{-1/2}B\right)\\ &=\operatorname{tr}\left(2X-2Z-X^{-1}Y^2\right)\\ &=\operatorname{tr}\left(2X-2Z-X^{-1}(X^2-Z^2)\right)\\ &=\operatorname{tr}\left(X-2Z+X^{-1}Z^2\right)\\ &=\left\|X^{1/2}-X^{-1/2}Z\right\|_F^2\ge0. \end{aligned}