Let $A,B\in\mathfrak{A}$ be two positive elements s.t $A\geq B\geq 0$. Then $A^{\frac{1}{2}}\geq B^{\frac{1}{2}}$.
This is an exericse from "$C^*$ algebras by example" by Kenneth Davidson, and as a hint he says that we need to use the fact that if $A,B\in\mathfrak{A}_{sa}$ are invertible then $A\geq B\Rightarrow B^{-1}\geq A^{-1}$ to show that $\|A^{\frac{1}{2}}B^{-\frac{1}{2}}\|\leq 1$ and use another exercise to conclude that $spr(B^{-\frac{1}{4}}A^{\frac{1}{2}}B^{-\frac{1}{4}})\leq 1$.
First of all, I'm not sure I understand why $B^{-\frac{1}{2}}$ even exists, when the exercise does not assume invertebility of neither operators. Secondly even if they are invertibles, how does the hint help me?
I tried approaching this in a different manner - we know that we have a faithful representation $\pi:\mathfrak{A}\to B(\mathcal{H})$ for some Hilbert space $\mathcal{H}$. Identifying $A,B$ with $\pi(A),\pi(B)$, we know that for every $h\in\mathcal{H}$: $$\langle Ah,h\rangle\geq\langle Bh,h\rangle$$ and therefore $$\langle A^{\frac{1}{2}}h,A^{\frac{1}{2}}h\rangle\geq\langle B^{\frac{1}{2}}h,B^{\frac{1}{2}}h\rangle$$ But it's not apparent to me if this implies that $A^{\frac{1}{2}}\geq B^{\frac{1}{2}}$ (it does imply that $\|A^{\frac{1}{2}}\|\geq\|B^\frac{1}{2}\|$ though).
Is my approach correct and if not how can I approach this using Davidson's hint? Any help would be appreciated.
I don't know an approach that doesn't use Davidson's argument, so I don't see how your attempt could be used.
As for the suggestion, while it is not said in the statement of the exercise, you are supposed to first assume that $B$ is invertible (just work with $B+\frac1n$). So you start with $A\leq B$, and multiply left and right with $B^{-1/2}$ (preserves order because you are conjugating with an element and its adjoint) to get $$ (A^{1/2}B^{-1/2})^*A^{1/2}B^{-1/2}=B^{-1/2}AB^{-1/2}\leq 1. $$