If $A^2\succeq B^2$ then $A\succeq B$

Question

Let $A$ and $B$ be $n×n$ symmetric matrices such that $A,B\succeq0$ ($A$ and $B$ are positive semidefinite) and $A^2\succeq B^2$. Is the following inequality true? $$A\succeq B$$ In this answer, it's shown that (I couldn't understand the proof) if we replace "semidefinite" with "definite" in the above proposition, the inequality is true but I don't know whether it holds in this case.

Answer 1

Per the "limiting argument" in the linked answer, it suffices to show that if $(A + tI) \succeq B$ for all $t > 0$, then $A \succeq B$.

Recall that a symmetric real matrix $M$ is positive semidefinite if for all vectors $x \in \Bbb R^n$, $x^TMx \geq 0$. Let $t_k = 1/k$. We note that for each $k$, $(A + t_k I) - B$ is positive semidefinite, so that $x^T[(A + t_k I) - B]x \geq 0$. Now, we have $$ x^T(A - B)x = \lim_{k \to \infty}x^T[(A + t_k I) - B]x. $$ The limit of a sequence of non-negative numbers is necessarily non-negative. Thus, $x^T(A - B)x \geq 0$. Since this holds for all vectors $x$, we conclude that $A - B \succeq 0$, which is to say that $A \succeq B$ as was desired.