Norms involving positive operators Let's say we have $A \leq B$. Is it then true that $||Ax|| \leq ||Bx||$ (where $x, A, B$ all belong to the same finite-dimensional Hilbert space $H$)?
 A: Counterexample:
$$
A=\left(\matrix{1&1\\1&1}\right)\qquad B=\left(\matrix{2&1\\1&1}\right)
$$
Details: Your question needs some context. In operator theory, this reads: is the function $f(t)=t^2$ operator monotone on $[0,+\infty)$, i.e does $A\leq B$ imply $f(A)\leq f(B)$ for every self-adjoint matrices with nonnegative spectrum? For instance, $g(t)=\sqrt{t}$ is operator monotone on $[0,+\infty)$. But $f(t)=t^2$ is not, as the above example shows.
Positive operators: An operator $A$ in $B(H)$ is called positive if and only if one of the following equivalent assertions holds. In this case, one writes $A\geq 0$.
1) $A$ is self-adjoint and $(Ax,x)\geq 0$ for all $x\in H$ (note that the latter implies the former in the complex case by polarization, but not in the real case).
2) $A$ is self-adjoint with nonnegative spectrum.
3) There exists $B$ self-adjoint such that $A=B^2$.
Partial order: For every self-adjoint operators $A,B$, one writes $A\leq B$ if $B-A\geq 0$. This defines a partial order. You question may be restated as follows:
$$
0\leq A\leq B\quad\Rightarrow?\quad A^2\leq B^2.
$$
Indeed, 
$$
\|Bx\|^2-\|Ax\|^2=(Bx,Bx)-(Ax,Ax)=(B^2x,x)-(A^2x,x)=((B^2-A^2)x,x).
$$
Counterexample check: It is easily seen that $A$ and $B$ are positive, as they are self-adjoint with spectra $\{0,2\}$ and $\{\frac{3\pm\sqrt{5}}{2}\}$. Now $B-A\geq 0$ as it is self-adjoint with spectrum is $\{0,1\}$. But $B^2-A^2$ has determinant $-1$, so it is not positive.
Commuting case: If we assume that $A$ and $B$ commute, the result holds. With matrices, you can prove it easily by simultaneous diagonalization in an orthonormal basis. I will now give a proof which works in any dimension. 
Write $B^2-A^2=(B-A)(A+B)$. Then observe that $B-A$ and $A+B$ are two commuting positive operators. Write $B-A=C^2$ and $A+B=D^2$ with $C=\sqrt{B-A}$ and $D=\sqrt{A+B}$ the unique positive square roots given by continuous functional calculus for normal operators. Since $B-A$ and $A+B$ commute, $C$ and $D$ commute by functional calculus. Therefore $B^2-A^2=C^2D^2=(CD)^2\geq 0$.
