1<(3+√2)/(2+√3)<2
[tex]\it 1\ \textless \ \dfrac{3+\sqrt2}{2+\sqrt3}\ \textless \ 2 \Leftrightarrow 2 + \sqrt3 \ \textless \ 3+ \sqrt2 \ \textless \ 4+2\sqrt3[/tex]
Avem o dublă inegalitate, prin urmare vom avea două cazuri:
[tex]\it I) \ \ 2+\sqrt3\ \textless \ 3+\sqrt2 \Leftrightarrow (2+\sqrt3)^2 \ \textless \ (3+\sqrt2)^2 \Leftrightarrow \\\;\\
\Leftrightarrow 5+ 4\sqrt3\ \textless \ 5+6\sqrt2 \Leftrightarrow 4\sqrt{3} \ \textless \ 6\sqrt{2} \Lefrightarrow \sqrt{4^2\cdot3} \ \textless \ \sqrt{6^2\cdot2} \Leftrightarrow
\\\;\\
\Leftrightarrow \sqrt{48} \ \textless \ \sqrt{72} \ \ (A)[/tex]
