$0\leq a \leq b $ implies $ b^{-1} \leq a^{-1}$ in C*-algebra? I'm studying C*-algebra in Conway's Functional Analysis.
Problem is this.

$0\leq a \leq b $, then $ b^{-1} \leq a^{-1}$ in C*-algebra?

I tried this problem via this way.
1st trial) Since $  a^{-1} - b^{-1} = a^{-1}(b-a)b^{-1}$ so product of three positive element is positive.
But this is false because product of positive element in C*-algebra is not positive.
2nd trial) How about using below Lemma?

$a \geq 0$  is equivalent to $a= a^*, ||t-a||\leq t$ for some $t \geq ||a|| $

Then, since $||a^{-1}(b-a)b^{-1}||  \leq ||a^{-1}|| \cdot ||b-a|| \cdot ||b^{-1}||$ so set $t = ||a^{-1}|| \cdot ||b-a|| \cdot ||b^{-1}||$ and try to solve this problem. What I stuck is is $||a||^{-1} = ||a^{-1}||$? It is just kind of passing thought, but in generally $a$ is hermitian, then $||a^{2^n}|| = ||a||^{2^n}$ so i think this might be true.
Can you help me?
 A: One way to prove the result is to first realise the $C^*$-algebra as sub-algebra of $\mathcal{B}(H)$ for a Hilbert space $H$ (GNS-construction) and then use the Lemmata below.
Lemma 1 Suppose $H$ is a Hilbert space and $S,T\in \mathcal{B}(H)$ are invertible, self-adjoint operators. Then $\Vert Sx\Vert \le \Vert T x \Vert$ for all $x\in H$ implies $\Vert T^{-1}x\Vert\le \Vert S^{-1} x\Vert$ for all $x\in H$.
Proof. Let $x,y\in H$, then
$$
\vert \langle T^{-1}x,y\rangle \vert = \vert \langle S^{-1}x,ST^{-1}y\rangle\vert\le \Vert S^{-1}x\Vert \cdot \Vert ST^{-1}y\Vert \le \Vert S^{-1}x\Vert \cdot\Vert y \Vert.
$$
Here we first used the Cauchy-Schwarz inequality and then the assumed estimate $\Vert Sz\Vert \le \Vert Tz \Vert$ for $z=T^{-1}y.$  The Lemma follows from taking the supremum over all $y\in H$ with $\Vert y \Vert = 1$.
Lemma 2 Suppose $H$ is a Hilbert space and $A,B\in \mathcal{B}(H)$ are invertible, self-adjoint operators. Then $A\le B$ implies $B^{-1}\le A^{-1}$.
Proof. Apply the previous Lemma for $S=A^{1/2}$ and $T=B^{1/2}$.
I've stolen the trick above from
a paper by Kato, where he shows a similar result for unbounded operators. I.e. when one defines $A\le B$ correctly (paying attention to domains), then $0<A\le B$ implies $0<B^{-1}\le A^{-1}$. This can be used to derive operator monotonicity of the maps $f(z)=z^\nu$ ($0\le \nu \le 1$), i.e. $0\le A \le B$ implies $f(A)\le f(B)$.
One more remark: It is not true in general that $\Vert a^{-1} \Vert = \Vert a \Vert^{-1}$. To see this, assume that $a_n$ is a sequence of invertible elements with limit $a_*\neq 0$. Then the sequence $\Vert a_n^{-1} \Vert = \Vert a_n \Vert^{-1}$ would be bounded by some $C<\infty$. Hence $\Vert a_n^{-1} - a_m^{-1}\Vert \le C^2 \Vert a_n - a_m \Vert \rightarrow 0$, which means that $a_n^{-1}$ must have a limit and it's easy to see that this must be an inverse of $a_*$. But this would imply that the set of invertible elements is open and closed in the set of nonzero elements and thus every nonzero $a$ was invertible, which is absurd unless the $C^*$-algebra is $\mathbb{C}$.
A: You may also note that if $a,b$ are invertible and
$$b-a ≥0$$
then:
$$a^{-1/2}(b-a)a^{-1/2}= a^{-1/2}ba^{-1/2}- 1≥0$$
and $a^{-1/2}ba^{-1/2}≥1$. Since $1$ commutes with that element you may take the inverse of both sides, turning the inequality around to get:
$$1-a^{1/2}b^{-1}a^{1/2}≥0$$
now conjugate with $a^{-1/2}$ again to get
$$a^{-1}-b^{-1}≥0$$
