Is $\sqrt{1+x^2}$ matrix monotone? A function $f(x)$ is matrix monotone if $f(A)-f(B)$ is positive semidefinite whenever $A-B$ is positive semidefinite for positive semidefinite matrices $A, B$. Is $\sqrt{1+x^2}$ matrix monotone?
No. One may be tempted to expect $\sqrt{1+x^2}$ to behave like $x$. However, since $x^2$ is not matrix monotone, it may cause $\sqrt{1+x^2}$ to fail to be matrix monotone. I don't have a nice counterexample at hand, but if you have access to a numerical linear algebra package, you will see that when
$$
A=\pmatrix{2.01&amp;1\\ 1&amp;3},\ B=\pmatrix{1\\ &amp;2},
$$
we have $A\succ B\succ 0$ but the eigenvalues of $\sqrt{I+A^2}-\sqrt{I+B^2}$ are $1.83$ and $-8.34\times10^{-3}$.
I think yes.  My general thinking is that the positive definite property is tied
to a matrix's determinate being non-zero, and that the square of a matrix will merely
amplify the magnitude of the determinate, but not zero it out.

I emphasize that this is strictly a guess.