Let $A,B$ be positive semidefinite self-adjoint operator on some finite inner product space. Is it true that $|Ax|\leq |Bx|$ for every vector $x$ iff $A\leq B$ [this notation means $B-A$ is positive semidefinite, i.e. $(Ax,x)\leq (Bx,x)$ for every $x$]? In this case, if $\text{tr}A=\text{tr}B$ holds, is it true that $A=B$?
I can see that using the square root, $A\leq B$ iff $(A^{1/2}x,A^{1/2}x)\leq (B^{1/2}x,B^{1/2}x)$, so indeed $|Ax|\leq |Bx|$; in this case if $\text{tr}A=\text{tr}B$ but $B-A$ is both traceless and positive semidefinite so $A=B$. What if we only know $|Ax|\leq |Bx|$, this is equivalent to ask if the partial order $"\leq"$ satisfies $A^2\leq B^2$ implies $A\leq B$?
