$(a,b)R(a,b)$ is always true , since $a<=a \ and\ b <= b$ is always true , thus the relation is reflexive for sure.
According to the definition of transitivity :-
$If \ (a,b)R(c,d) \ and \ (c,d)R(e,f) \ then (a,b)R(e,f) \ must \ also \ hold \ true.$
Here , the statement is in the form of an implication ,
If A then B.
$(\ (a,b)R(c,d) \ and \ (c,d)R(e,f) )\ => (a,b)R(e,f) $
If the RHS can be proved to be false and LHS can be proved to be true at the same point of time we can surely say that the statement is invalid and thus the relation in not transitive.
Let RHS is false.
Thus , $((a,b),(e,f)) \notin R => a>e \ and \ b>f.$
Let's reverse engineer this .
Let $a = 4 , b=5 ,e=3 \ and \ f=3$.
It can be easily shown that:-
$(4,5)R(3,6) \ holds \ and \ (3,6)R(3,3) \ holds \ but \ (4,5)R(3,3) \ doesn't \ hold. $
Thus we can show that the LHS of the implication is true , and RHS is false at the same point of time , thus the statement is invalid.
Thus , the relation is not following the property of transitivity.