If $A\ B$ are symmetric positive definite matrices then $A>B$ iff $\sqrt{A}>\sqrt{B}$ [duplicate]

My guess is, it only holds one way i.e. $A>B$ then $\sqrt{A}>\sqrt{B}$ but not otherwise. Any proof or counterexample would be appreciated.
• $A>B$ refer to partial ordering of SPD matrices. It implies $A-B$ is positive definite – piyush_sao Oct 16 '13 at 4:56