if the matrix such $B-A,A$ is Positive-semidefinite,then $\sqrt{B}-\sqrt{A}$ is Positive-semidefinite Question:
let the matrix $A,B$ such  $B-A,A$ is Positive-semidefinite
show that:

$\sqrt{B}-\sqrt{A}$ is Positive-semidefinite

maybe The general is true?
question 2:
(2)$\sqrt[k]{B}-\sqrt[k]{A}$ is Positive-semidefinite
This problem is very nice,because we are all know this
if $$x\ge y\ge 0$$,then we have
$$\sqrt{x}\ge \sqrt{y}$$
But in matrix,then this is also true,But I can't prove it.Thank you

Another proof (short and simple) from "Linear Algebra and Linear Models" by R. B. Bapat. 
Lemma Let $A$ and $B$ be $n\times n$ symmetric matrices such that $A$ is positive definite and $AB+BA$ is positive semidefinite, then Y is positive semidefinite. 
Proof of $B\geq A \implies B^{\frac{1}{2}}\geq A^{\frac{1}{2}}$
First consider the case, when $A$ and $B$ are positive definite. 
Let $X=(B^{\frac{1}{2}}+ A^{\frac{1}{2}})$ and $ Y=(B^{\frac{1}{2}}- A^{\frac{1}{2}})$, 
then $XY+YX=2(B-A)$ 
Now, $(B-A)$ is positive semidefinite implies (given) $\implies 2(B-A)$ is positive semidefinite. Also $X=(B^{\frac{1}{2}}+ A^{\frac{1}{2}})$ is positive definite as positive linear combination of positive definite matrices is positive definite.
Hence by the lemma, $Y=(B^{\frac{1}{2}}- A^{\frac{1}{2}})$ is positive semidefinite. Therefore, $B^{\frac{1}{2}}\geq A^{\frac{1}{2}}$
The case, when $A$ and $B$ are positive semidefinite matrices can be dealt as the other answer.
