On the square root function of matrices Let $A, B$ be positive definite matrices and let $P$ be an orthogonal projection. If $A \leq PBP,$ does it follow that
 $$ A^{1/2} \leq PB^{1/2}P?$$
 A: If $A,B\geq 0$, then $A\leq B \implies \sqrt{A}\leq \sqrt{B}$. There is a great proof by Kato but I don't remember it. 
Yet,  $A\leq B$ doesn't imply $A^2\leq B^2$.
EDIT. About the zoli's question, clearly the answer is yes when $P=I$. Yet, when $rank(P)<n$, then necessarily $rank(A)<n$, that is, the problem is poorly stated; we must assume that $A,B\geq 0$ (note that $(PBP)^{1/2}\not= PB^{1/2}P$).
A: Since $0\lt A\le PBP$, $P$ must be nonsingular, but then it has to be the identity matrix. So the inequality boils down to $A^{1/2}\le B^{1/2}$, which is true because the square root function is operator monotone. Here is the usual proof: $0\lt A\le B$ implies that $B^{-1/2}AB^{-1/2}\leq I$. Hence
$$
\rho(B^{-1/4}A^{1/2}B^{-1/4})=\rho(B^{-1/2}A^{1/2})\le\sigma(B^{-1/2}A^{1/2})=\rho(B^{-1/2}AB^{-1/2})^{1/2}\leq 1
$$
and in turn $B^{-1/4}A^{1/2}B^{-1/4}\le I$, i.e. $A^{1/2}\le B^{1/2}$.
Edit. The assumption that $A$ is positive definite is essential. If $A$ is merely positive semidefinte, the inequality would be false. Here is a simple counterexample:
$$
A=A^{1/2}=P=\pmatrix{1&0\\ 0&0},\ B=\pmatrix{1&1\\ 1&2},\ B^{1/2}=\frac1{\sqrt{5}}\pmatrix{2&1\\ 1&3}.
$$
Note that the usual continuity argument doesn't work because the jump from $P=I$ to a singular orthogonal projection is not continuous.
