TITLE: Showing $\sqrt{2}\sqrt{3} $ is greater or less than $ \sqrt{2} + \sqrt{3} $ algebraically
QUESTION [15 upvotes]: 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?

REPLY [5 votes]: Another very straightforward way which works for many simple inequalities is to show that the inequality is logically equivalent to a tautology:
As we are only dealing with positive numbers, we have
$$ \sqrt 2 + \sqrt 3 > \sqrt 2 \sqrt 3 \\ \Leftrightarrow \;(\sqrt 2 + \sqrt 3)^2 > 6 \\ \Leftrightarrow\; 2\sqrt 6 > 1.$$
The last inequality holds since $\sqrt x \geq 1$ for all $x\geq 1$.