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
 A: Two proofs: (thanks for Julien for helping me fix the first proof)
First proof: 
First assume $A,B>0$. Suppose $B \geq A$.
$$B-A \geq 0 \\ A^{-1/2}BA^{-1/2} - I \geq 0, $$
since $M\geq 0 \Rightarrow N^{*}MN \geq 0$ for $N$ invertible. This means all eigenvalues of $A^{-1/2}BA^{-1/2}$ are strictly greater than 1. Therefore for all $\|x\|=1$
$$\langle A^{-1/2}BA^{-1/2}x,x \rangle \geq 1 \\ \langle BA^{-1/2}x, A^{-1/2}x \rangle \geq 1 \\ \langle B^{1/2}A^{-1/2}x, B^{1/2}A^{-1/2}x \rangle = \|B^{1/2}A^{-1/2} x\|^2\geq1 \\ \|B^{1/2}A^{-1/2} x\|\geq1$$
so all eigenvalues of $B^{1/2}A^{-1/2}$ are strictly greater than 1 in modulus. But by a similarity transformation, these are also the eigenvalues of $A^{-1/4}B^{1/2}A^{-1/4}$, and those are all positive since $A^{-1/4}B^{1/2}A^{-1/4}$ is positive (being of the form $N^* M N$ for $M>0$ and $N$ invertible). So 
$$A^{-1/4}B^{1/2}A^{-1/4}-I \geq0\\B^{1/2}-A^{1/2}\geq0,$$
finishing the case $A,B>0$. To extend to $A,B\geq 0$, let $A_\epsilon:=A+\epsilon I$, and similarly for $B_\epsilon$. Then $B\geq A \Rightarrow B_\epsilon \geq A_\epsilon \Rightarrow (B_\epsilon)^{1/2}\geq (A_{\epsilon})^{1/2} \Rightarrow B^{1/2}\geq A^{1/2}$, if you believe that $\epsilon \mapsto M_\epsilon^{1/2}-N_{\epsilon}^{1/2}$ is right-continuous in the eigenvalues at zero (follows from $\epsilon \mapsto M_\epsilon^{1/2}$ being right-continuous in the eigenvalues).
Second Proof (from Lax's book Linear Algebra and its Applications): This relies on the fact that if $A(t)$ is a matrix-valued function whose derivative is everywhere positive definite, then $t_2 > t_1 \Rightarrow A(t_2) > A(t_1)$ (unstrict inequality corresponds to positive semidefiniteness). Also it uses the lemma that if $A$ is positive definite and $AB+BA$ is positive definite, then $B$ is positive definite. And it uses that positive semidefinite matrices are a convex subset of $\mathbb{C}^{n^2}$ (all these proofs can be found in Lax's book in the section on matrix inequalities).
Let $B\geq A \geq 0$. Define a function $M(t) = A + t(B-A)$. It is positive on $[0,1]$ by convexity (as a matter of fact, it is positive for all $t\geq0$, I believe). Further, it has a positive semidefinite matrix as its derivative. Also $\sqrt{M(t)}$ is positive semidefinite. Define $R(t) = \sqrt{M(t)}$. Then $R^2 = M$, so 
$$R\dot{R} + \dot{R} R = M'(t) \geq 0.$$
$R$ is positive and $R\dot{R} + \dot{R} R$ is positive, so $\dot{R}$ is positive. Therefore $R$ is a non-decreasing function of $t$ and $R(1) \geq R(0)$. QED
A: 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.
