$(\sqrt{2}\sqrt{3})^2=6<5+2<5+2\sqrt{6}=2+3+2\sqrt{2}\sqrt{3}=(\sqrt{2}+\sqrt{3})^2$, so $\sqrt{2}\sqrt{3}<\sqrt{2}+\sqrt{3}$.

Alternatively, $\sqrt{2}+\sqrt{3}>\sqrt{2}+\sqrt{2}=2\sqrt{2}>\sqrt{3}\sqrt{2}$.

Method 1: $\sqrt{2}+\sqrt{3}>\sqrt{2}+\sqrt{2}=2\sqrt{2}>\sqrt{3}\sqrt{2}$.

Method 2: $(\sqrt{2}\sqrt{3})^2=6<5+2<5+2\sqrt{6}=2+3+2\sqrt{2}\sqrt{3}=(\sqrt{2}+\sqrt{3})^2$, so $\sqrt{2}\sqrt{3}<\sqrt{2}+\sqrt{3}$.

Method 3: $\frac{196}{100}<2<\frac{225}{100}$ and $\frac{289}{100}<3<\frac{324}{100}$, so $\sqrt{2}\sqrt{3}<\frac{15}{10}\frac{18}{10}=\frac{270}{100}<\frac{14}{10}+\frac{17}{10}<\sqrt{2}+\sqrt{3}$.

$(\sqrt{2}\sqrt{3})^2=6<5+2<5+2\sqrt{6}=2+3+2\sqrt{2}\sqrt{3}=(\sqrt{2}+\sqrt{3})^2$, so $\sqrt{2}\sqrt{3}<\sqrt{2}+\sqrt{3}$.

$(\sqrt{2}\sqrt{3})^2=6<5+2<5+2\sqrt{6}=2+3+2\sqrt{2}\sqrt{3}=(\sqrt{2}+\sqrt{3})^2$, so $\sqrt{2}\sqrt{3}<\sqrt{2}+\sqrt{3}$.

Alternatively, $\sqrt{2}+\sqrt{3}>\sqrt{2}+\sqrt{2}=2\sqrt{2}>\sqrt{3}\sqrt{2}$.