assume we have $A,B,A-B$ non-negative definite matrix…

Question

Assume we have real symmetric $A,B,A-B$ non-negative definite matrix , how to prove $\sqrt{A}-\sqrt{B}$ also is non-negative definite matrix? Where $\sqrt{A}$ is the only symmetric non-negative definite matrix satisfy $X^2=A$.

Answer 1

Note that $\sqrt{A},\sqrt{B},$ and $\sqrt{A}^2 - \sqrt{B}^2$ are positive semidefinite. Then, apply the arguments from here.