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).

$6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$
$3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$
This implies that $\dfrac 52$ is between the two quantities:
\begin{align}
   \dfrac{49}{4} &> 12 \\
   \dfrac 72 &> 2\sqrt 3 \\
   \dfrac{12}{2} &> 2\sqrt 3 + \dfrac 52 \\
   6 - 2\sqrt 3 &> \dfrac 52
\end{align}
and
\begin{align}
   \dfrac{81}{4} &> 18 \\
   \dfrac 92 &> 3\sqrt 2 \\
   \dfrac 52 &> 3\sqrt 2 - 2
\end{align}
It follows that $6-2√3 > 3\sqrt 2 - 2$.
