In this post all my matrices will be $\mathbb R^{N\times N}$ symmetric positive semi-definite (psd), but I am also interested in the Hermitian case. In particular the square root $A^{\frac 12}$ of a psd matrix $A$ is defined unambigusouly via the spectral theorem. Also, I use the conventional Frobenius scalar product and norm $$ <A,B>:=Tr(A^tB), \qquad |A|^2:=<A,A> $$

Question: is the folowing inequality true $$ |A^{\frac 12}-B^{\frac 12}|^2\leq C_N |A-B|\quad ??? $$ for all psd matrices $A,B$ and a positive constant $C_N$ depending on the dimension only.

For non-negative scalar number (i-e $N=1$) this amounts to asking whether $|\sqrt a-\sqrt b|^2\leq C|a-b|$, which of course is true due to $|\sqrt a-\sqrt b|^2=|\sqrt a-\sqrt b|\times |\sqrt a-\sqrt b|\leq |\sqrt a-\sqrt b| \times |\sqrt a+\sqrt b|=|a-b|$.

If $A$ and $B$ commute then by simultaneous diagonalisation we can assume that $A=diag(a_i)$ and $B=diag(b_i)$, hence from the scalar case $$ |A^\frac 12-B^\frac 12|^2 =\sum\limits_{i=1}^N |\sqrt a_i-\sqrt b_i|^2 \leq \sum\limits_{i=1}^N |a_i-b_i| \leq \sqrt N \left(\sum\limits_{i=1}^N |a_i-b_i|^2\right)^\frac 12=\sqrt N |A-B| $$

Some hidden convexity seems to be involved, but in the general (non diagonal) case I am embarrasingly not even sure that the statement holds true and I cannot even get started. Since I am pretty sure that this is either blatantly false, or otherwise well-known and referenced, I would like to avoid wasting more time reinventing the wheel than I already have.

This post and that post seem to be related but do not quite get me where I want (unless I missed something?)

**Context**: this question arises for technical purposes in a problem I'm currently working on, related to the Bures distance between psd matrices, defined as
$$
d(A,B)=\min\limits_U |A^\frac 12-B^\frac 12U|
$$
(the infimum runs over unitary matrices $UU^t=Id$)