I asked a question earlier regarding self-adjoint elements. I will ask a similar question again.
Claim: I have $0\leq A_1\leq A_2$ for self-adjoint elements $A_1,A_2$ in $C^\ast$-algebra. If $0\leq A_1\leq A_2$ and where $A_1$ and $A_2$ commute, then we have $A_1^k\leq A_2^k$ for every positive integer $k$.
Idea: Can we show it by considering the following, i.e. if there are positive elements $B_j$ for $1\leq j \leq m$ with $$0\leq A_1\leq B_1\leq B_2\leq \cdots \leq B_m\leq A_2$$ so that any two neighbors in this list above commute, then $A^k\leq B^k$ for any positive integer $k$. How would one attack this? If we can’t how would we else show the claim?
I can't follow your idea but here is a proof.
The two positive elements $A:=A_1$ and $B:=A_2-A_1$ commute hence $$A_2^k-A_1^k=\sum_{j=1}^k\binom kjA^{j/2}B^{k-j}A^{j/2}\ge0.$$