Suppose we have symmetric matrices $A$ and $B$, where
$$A \succ 0, \qquad B \succeq 0, \qquad A - B \succeq 0$$
Here, $A \succeq B$ means that $x^\top Ax \geq x^\top Bx$ for all $x$.
Do we have $A^2\succeq B^2$?
If $S$ is positive definite, do we have $A S A \succeq B S B$?
It seems right, but I do not know how to prove or disprove it.
Question 1
No, for a counterexample consider $$ A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} $$ $$ B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$
Then $A-B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \geq 0$ but $A^2 - B^2 = \begin{pmatrix} 3 & 1 \\ 1 & 0\end{pmatrix}$ is not PSD.
Question 2
No, take $S$ to be the identity matrix then $S$ is positive definite but by part $1$ the inequality cannot hold.