$|A\times B|= \text{max}(|A|,|B|)$ for infinite sets I am fairly sure, given examples $\Bbb{R}\times \Bbb{R},\Bbb{R}\times \Bbb{Q},\Bbb{Q}\times \Bbb{Q} $, that this is correct, but do not know how to prove it. 
In my cited examples the proof has always used some specific property of $A,B$, leaving me clueless as to the general method of solving this. I am especially interested in the case were $|\Bbb{R}|>A>|\Bbb{Q}|,|B|=|\Bbb{Q}|$, though ideally the answer would be more general.
To be explicit, I'm asking for a proof of the statement in the question title, where $A,B$ are sets with infinite cardinalities.
 A: This requires the axiom of choice in all accounts.
First because the axiom of choice is equivalent to the fact that every two cardinals can be compared (otherwise $\max$ has no meaning); and secondly because we use the fact that $|A|^2=|A|$ for infinite sets, which is also equivalent to the axiom of choice.
Now, using these two facts we have that if $|B|\leq|A|$ then:
$$|A|\leq |A\times B|\leq |A\times A|\leq|A|$$
In either case you can't quite prove this "explicitly" without the axiom of choice. Even if you do know that $|A|<|B|$, it might still be the case that $|B|<|A\times B|$.
Related threads:


*

*Godel's pairing function and proving c = c*c for aleph cardinals

*For all infinite cardinals $\kappa, \ (\kappa \times \kappa, <_{cw}) \cong (\kappa, \in).$

*Can an infinite cardinal number be a sum of two smaller cardinal number? (This might talk about sums, but the same argument shows that defining multiplication by taking the maximal of the two implies choice.)

*For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$

*[To be added... maybe?]

A: Let $A$ be the biggest of the two
$$
|A|\le|A\times B|\le|A^2|
$$
The last piece you miss is $|A|=|A^2|$, that is equivalent to Choice Axiom, and can be proved, for example, through  Zorn's Lemma
A: Hint:$A\times B=\cup \{f:\{0,1\}\to A\cup B:f(0)\in A,f(1)\in B\}$.
Now suppose $|A|>|B|$ : 
$|A|=|A\times A|$ and $A\times A=\cup \{f:\{0,1\}\to A:f(0),f(1)\in A\}$.
Find a bijection between $A\times B$ and $A\times A$.
