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 $x < y \text{ or } x>y$.

A visual demonstration of sorts:
Starting from $ \ x \ = \ \sqrt3 - 1 \ \ , \ $ we have
$$ x \ + \ 1 \ \ = \ \ \sqrt3 \ \ \Rightarrow \ \ x^2 \ + \ 2x \ + \ 1 \ \ = \ \ 3 \ \ . \ $$   So $ \ \sqrt3 - 1 \ $ is the larger of the two zeroes of $ \ x^2 + 2x - 2 \ \ . \ $  As for $ \ x \ = \ \sqrt5 \ - \ \sqrt2 \ \ , \ $ we may write
$$ x \ + \ \sqrt2 \ \ = \ \ \sqrt5 \ \ \Rightarrow \ \ x^2 \ + \ 2 \sqrt2 · x  \ + \ 2 \ \ = \ \ 5  \ \ , $$
making $ \ \sqrt5 \ - \ \sqrt2 \ $ the larger of the two zeroes of the polynomial $ \ x^2   +   2 \sqrt2 · x    -   3 \ \ . \ $   A graph of the two polynomials (the first in red, the second in blue) indicates the relative positions of these two zeroes.
If one mistrusts a graph (a not unreasonable attitude), we can show that the curve for $ \ x^2 + 2x - 2 \ $ lies "above" that of $ \ x^2   +   2 \sqrt2 · x    -   3 \ \ $ [vertex at $ \ (-1 \ , \ -3) \ $ versus $ \ (-\sqrt2 \ , \ -5)  \ \ , \ $ "higher" $ \ y-$ intercept; also, $$   x^2 + 2x - 2 \ > \ x^2 + 2 \sqrt2 x - 3 \ \ \Rightarrow \ \ 1 \ > \ 2·(\sqrt2 - 1)·x $$ $$  \Rightarrow \ \ \frac{1}{2·(\sqrt2 - 1)} \ = \ \frac12·(1 + \sqrt2) \ > \ x \ \ . \ ] $$ until the curves intersect at $ \ x \ = \ \frac{1}{2 \sqrt2 \ - \ 1} \ = \ \frac12   +  \frac{\sqrt2}{2} \ \ , $ $  y \ = \ \frac32 \sqrt2 - \frac14 \ > \ 0 \ \ . \ $  Since this intersection is "above" the $ \ x-$axis, the positive $ \ x-$intercept of $ \ x^2 + 2x - 2 \ $ is smaller than that of $ \ x^2   +   2 \sqrt2 · x    -   3 \ \ . $

