$6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$
$3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$
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 $6-2√3 > 3\sqrt 2 - 2$.
<<<<< added 11/23/2043 >>>>>
What I am looking for is a simple, improvable, method for
finding nice bounds for $6-2√3$ and $3\sqrt 2 - 2$.
First pass:
$$ \left( 2\sqrt 3 \right)^2 = 12 $$
$$ 9 < 12 < 16 $$
$$ 3^2 < \left( 2\sqrt 3 \right)^2 < 4^2 $$
$$ 3 < 2\sqrt 3 < 4 $$
$$ -4 < -2\sqrt 3 < -3 $$
$$ 2 < 6 - 2\sqrt 3 < 3$$
and
$$ \left( 3 \sqrt 2 \right)^2 = 18 $$
$$ 16 < 18 < 25 $$
$$ 4^2 < \left( 3 \sqrt 2 \right)^2 < 5^2 $$
$$ 4 < 3 \sqrt 2 < 5 $$
$$ 2 < 3 \sqrt 2 - 2 < 3 $$
And all we have accomplished is to show that both quantities are between 2 and 3. So we double the two quantities; that is, we compare
$12-4√3 $ and $6√2-4$
$$ \left( 4\sqrt 3 \right)^2 = 48 $$
$$ 36 < 48 < 49 $$
$$ 6^2 < \left( 4 \sqrt 3 \right)^2 < 7^2 $$
$$ 6 < 4\sqrt 3 < 7 $$
$$ -7 < -4\sqrt 3 < -6 $$
$$ 5 < 12 - 4\sqrt 3 < 6 $$
$$ \dfrac 52 < 6 - 2\sqrt 3 < 3 $$
and
$$ \left( 6 \sqrt 2 \right)^2 = 72 $$
$$ 64 < 72 < 81 $$
$$ 8^2 < \left( 6 \sqrt 2 \right)^2 < 9^2 $$
$$ 8 < 6 \sqrt 2 < 9 $$
$$ 4 < 6 \sqrt 2 - 4 < 5 $$
$$ 2 < 3 \sqrt 2 - 2 < \dfrac 52 $$
And we conclude that
$$ 2 < 3 \sqrt 2 - 2 < \dfrac 52 < 6 - 2\sqrt 3 < 3 $$