Maximum of two positive operators Let $A,B$ be two positive operators in $B(H)$. Does there exist, in general, an operator $C$ such that for each $T$, if $A \leq T$ and $B \leq T$, then $$A\leq C \leq T\quad  \text{and}\quad B\leq C\leq T?$$
That is, does there exist the maximum operator of any two positive operators?
 A: Such maximum does not exist in general. Consider the following example (I didn't work very hard to check whether a simpler example is available). Let 
$$
A=\begin{bmatrix}1&0\\0&2\end{bmatrix},\ \ B=\begin{bmatrix}3&1\\1&1\end{bmatrix}
$$
and consider 
$$
X=\begin{bmatrix}3&1\\1&5/2\end{bmatrix}, \ \ Y=\begin{bmatrix}7/2&1/2\\1/2&5/2\end{bmatrix}.
$$
Then $$A\leq X\leq A+B,$$ $$B\leq X\leq A+B,$$ $$A\leq Y\leq A+B,$$ $$B\leq Y\leq A+B.$$
But $X$ and $Y$ are not comparable. We can use this to show there is not $C$ above $A,B$ and below $X,Y$.
Indeed, assume that there exists $C$ with $B\leq C\leq X$. Such $C$ is of the form 
$$
C=\begin{bmatrix}3&1\\1&z\end{bmatrix},
$$
with $1\leq z\leq 5/2$. To get $A\leq C$, we need $z\geq5/2$, so $z=5/2$. But then $C\not\leq Y$: we have
$$
Y-C=\begin{bmatrix}1/2&-1/2\\-1/2&0\end{bmatrix},
$$
which is not positive. 
A: There is a classical conclusion that if A and B are positive in B(H), and max{||A||,||B||}<1, then there exists some positive C such that A, B$\leq$ C.
Do you want to fine this conclusion?
