Suppose we take $A=B(H)$,where $H$ is a complex Hilbert space.Let $S,T$ be two positive invertible elements in $A$,$K=\{m>0,S\leq m T\}$, when $dim(H)=1$,it is easy to see that $K$ has a minimum,when $dim(H)=n$,$S,T$ can be viewed as two $n\times n$ postive invertible real matrices.
If $dim(H)\geq 2$,does there must exist a minimal elment in $K$?
The set $K$ is always of the form $[r,\infty)$ for some $r\geq0$.
To see the above, let $r=\inf K$. For any $\varepsilon>0$, there exists $r_1\in K$ with $r_1<r+\varepsilon$. Then, for any vector $x\in H$, $$ \langle Sx,x\rangle\leq\langle r_1Tx,x\rangle\leq\langle (r+\varepsilon)Tx,x\rangle\leq\langle rTx,x\rangle+\varepsilon\langle Tx,x\rangle. $$ As this works for any $\varepsilon$, we get that $S\leq rT$.