[tex]\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}[/tex]
[tex](\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}\\(\sqrt{5}+1)^2=6+2\sqrt{5}=5+1+2\sqrt{5}>5+1+2\sqrt{4}=5+5=5+2.\frac{5}{2}=5+2\sqrt{\frac{25}{4}}>5+2\sqrt{\frac{24}{4}}=5+2\sqrt{6}\\\rightarrow (\sqrt{5}+1)^2>(\sqrt{3}+\sqrt{2})^2\\\rightarrow \sqrt{5}+1>\sqrt{3}+\sqrt{2}[/tex]