Showing $\sqrt{2}\sqrt{3} $ is greater or less than $ \sqrt{2} + \sqrt{3} $ algebraically How can we establish algebraically if $\sqrt{2}\sqrt{3}$ is greater than or less than $\sqrt{2} + \sqrt{3}$?
I know I can plug the values into any calculator and compare the digits, but that is not very satisfying. I've tried to solve $$\sqrt{2}+\sqrt{3}+x=\sqrt{2}\sqrt{3} $$ to see if $x$ is positive or negative. But I'm just getting sums of square roots whose positive or negative values are not obvious.
Can it be done without the decimal expansion?

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 $\sqrt{2}\sqrt{3}<\frac{15}{10}\frac{18}{10}=\frac{270}{100}<\frac{14}{10}+\frac{17}{10}<\sqrt{2}+\sqrt{3}$.
