For positive invertible operators $C\leq T$ on a Hilbert space, does it follow that $T^{-1}\leq C^{-1}$?

I need the following result. I think it's quite obvious but I don't know how to prove that: Let $C, T : \mathcal{H} \rightarrow \mathcal{H}$ be two positive, bounded, self-adjoint, invertible operators on a Hilbert space $\mathcal{H}$ such that $C \leq T$. Then it follows also that $T^{-1} \leq C^{-1}$. Or maybe this is even not true? If this is not true can somebody give me a counter-example or if it is true some strategy how to solve this? I would be very thankful.

mika

Solution 1:

Note that $T \ge C$ iff $C^{-1/2} T C^{-1/2} \ge I$ iff $C^{1/2} T^{-1} C^{1/2} = (C^{-1/2} T C^{-1/2})^{-1} \le I$ iff $T^{-1} \le C^{-1}$