<s>[INST] 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).
 [/INST] Here's yet another way, for those who aren't comfortable with the $\gtrless$ or $\sim$ notation.
We can use crude rational approximations to $\sqrt 2$ and $\sqrt 3$.
$$\begin{align}
\left(\frac{3}{2}\right)^2 = \frac{9}{4} & \gt 2\\
\frac{3}{2} & \gt \sqrt 2\\
\frac{9}{2} & \gt 3\sqrt 2
\end{align}$$
And
$$\begin{align}
\left(\frac{7}{4}\right)^2 = \frac{49}{16} & \gt 3\\
\frac{7}{4} & \gt \sqrt 3\\
\frac{7}{2} & \gt 2\sqrt 3
\end{align}$$
Adding those two approximations, we get
$$\begin{align}
\frac{9}{2} + \frac{7}{2} = 8 & \gt 3\sqrt 2 + 2\sqrt 3\\
6 + 2 & \gt 3\sqrt 2 + 2\sqrt 3\\
6 - 2\sqrt 3 & \gt 3\sqrt 2 - 2
\end{align}$$
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