If $0\leq A\leq B$ on Hilbert space and $A^{-1}$ exists, show that $A^{-1}\geq B^{-1}$ Does  anyone know how to show this? Let $H$ be a Hilbert space and $A$, $B$ bounded positive operators defined on $H$ such that $A^{-1}: H \rightarrow H$ exists and hence bounded and $A \leq B$. Therefore, $B^{-1}: H \rightarrow H$ exists and bounded. Show that $A^{-1} \geq B^{-1}$. 
I tried to use this theorem "If two bounded self-adjoint linear operators $S$ and $T$ on a Hilbert space $H$ are positive and commute, then their product $ST$ is positive." However, I don't know whether $A$ and $B$ commute with each other. Are there any other ways to deal with it? I am using book of Kreyszig.  

Since this is homework, I am only supposed to give hints.
$ A \le B \Leftrightarrow UAU^* \le UBU^* $ for any invertible $U$.
Hint 1: $ A \le B \Rightarrow B^{-1/2} A B^{-1/2} \le I $.
Hint 2: If $V$ is invertible, then $ VV^* \le I \Rightarrow V^*VV^*V \le V^* V \Rightarrow V^*V \le I$ (use $U = (V^*V)^{-1/2}$ for the last implication).
Hint 3: So $ B^{-1/2} A B^{-1/2} \le I \Rightarrow A^{1/2}B^{-1} A^{1/2} \le I $.
