Showing $\sqrt{2}\sqrt{3} $ is greater or less than $ \sqrt{2} + \sqrt{3} $ algebraically

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

$$\sqrt{3}\sqrt{2}-\sqrt{3}-\sqrt{2}+1=(\sqrt{3}-1)(\sqrt{2}-1) < 1$$

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.

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 $x\geq 1$.