Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).

Define
$a=6-2\sqrt 3>0$
$b=3\sqrt 2-2>0$
$a-b = 8 - (2\sqrt 3 + 3\sqrt 2)$
$(2\sqrt 3 + 3\sqrt 2)^2 = 30+12\sqrt 6 = 6×(5+2\sqrt 6)
 < 60 < 64$
because $6=2×3 < (5/2)^2$
$a-b > 8-8=0, a>b$
