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?

$$\sqrt{3}\sqrt{2}-\sqrt{3}-\sqrt{2}+1=(\sqrt{3}-1)(\sqrt{2}-1) < 1$$
The last inequality follows from the fact that $(\sqrt{3}-1)$ and $(\sqrt{2}-1)$ are in $(0,1)$.
Second solution
By AM-GM you have 
$$2\sqrt[4]{6} \leq  \sqrt{2}+\sqrt{3}$$
Combine this with $\sqrt[4]{6} < 2$, which is easy to prove, and you are done.
And a non-algebraic one, which is an overkill :)
Let $\theta$ be the angle so that $\cos(\theta)=-\frac{1}{2\sqrt{6}}$. Plot a point $A$ draw two rays with an angle of $\theta$ between them, and pick points $B$ respectively $C$ on these ray so that $AB=\sqrt{2}$ and $AC=\sqrt{3}$. By the cosine law
$$BC^2=2+3+2\sqrt{2}\sqrt{3}\frac{1}{2\sqrt{6}}=6$$
Thus the triangle $ABC$ has the edges of length $\sqrt{2},  \sqrt{3}$ and $\sqrt{6}$, and your inequality is exactly the triangle inequality.
