Is $Tr( (A^{-1}B)^2 ) \leq Tr ( A^{-1}B )$ if $A \geq B \geq 0$ Let $A$ be simmetric positive definite, $B$ simetric positive semi definite. And $A - B$ positive semi definite, i.e $A \geq B$
How can I convince myself that $$Tr( (A^{-1}B)^2 ) \leq Tr ( A^{-1}B )$$
 A: YES 
I found the answer to my question shortly after asking. I will post it so nayone else who has the same question can use it.
If all eigenvalues of $A^{-1/2}BA^{-1/2}$ $\lambda_1, \ldots \lambda_n$ are are smaller than $1$. Then  $Tr(  (A^{-1}B)^2  ) = Tr(  (A^{-1/2}BA^{-1/2})^2) = \sum \lambda_i^2 \leq \sum \lambda_i = Tr( A^{-1/2}BA^{-1/2} ) = Tr ( A^{-1}B )$.
Now notice that $A^{-1/2}BA^{-1/2} \leq Id$, because $ A \geq B \implies Id \geq A^{-1/2}BA^{-1/2}$. This gives that $0 \leq \lambda_i \leq 1$. 
A: Same solution was on the road here.
It's an "algebraic" solution that works in a general Hilbert space $H$ (with a trace on some operator algebra on it).
Let $\bar A$ be the inverse of $A$, $\bar A=A^{-1}$.
It is enough to show
$$
\underbrace{\bar A^{1/2}\; B\bar A B\; \bar A^{1/2}}
_{(\bar A^{1/2}\; B\; \bar A^{1/2})^2}
\le 
\bar A^{1/2}\; B\; \bar A^{1/2}\ ,
$$
since after applying the trace we can "move" the one $\bar A^{1/2}$ from one end to the other end. 
For the above, it is further successively enough to show the following inequalities involving selfadjoint operators $\ge 0$:
$$
\begin{aligned}
\bar A^{1/2}\; B\; \bar A^{1/2} &\le I \ ,\\
A^{1/2}\bar A^{1/2}\ B\; \bar A^{1/2} A^{1/2}&\le A^{1/2}\ I\ A^{1/2} \ ,\\
B&\le A\ .
\end{aligned}
$$
The last inequality is given.
