How to order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2}$ ascendingly? How would I order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $ without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.
What I tried to so far, because they are all positive numbers, is to square $x,y,z$ but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that $x<z$ is easy and so is $y<z$, but I'm stuck at $x < y \text{ or } x>y$.

Suppose $ x \geqslant y $, then
$$ \begin{array} { r c l }
\sqrt3 + \sqrt2 &\geqslant& \sqrt5 + 1 \\
(\sqrt3 + \sqrt2)^2 &\geqslant& (\sqrt5 + 1)^2 \\
5 + 2 \sqrt6 &\geqslant& 6 + 2 \sqrt5 \\
2( \sqrt6 - \sqrt5) &\geqslant& 1 \\
2( \sqrt6 - \sqrt5)( \sqrt6 + \sqrt5) &\geqslant& ( \sqrt6 + \sqrt5) \\
2(6 - 5) &\geqslant&  \sqrt6 + \sqrt5 \\
2 &\geqslant&  \sqrt6 + \sqrt5 > \sqrt4 + \sqrt4 = 4 \\
\end{array} 
$$
A contradiction.
