algebra precalculus - $sqrt{6}-sqrt{2}-sqrt{3}$ is irrational

Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational

My attempt:-
Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have

$$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$
Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$
and now squaring this we get $$6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$ . This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.

Answer

$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is a root of $x^4 - 22 x^2 - 48 x - 23$.

By the rational root theorem, $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is either irrational or an integer.

But

$$1.4 < \sqrt 2 < 1.5 \\ 1.7 < \sqrt 3 < 1.8 \\ 2.4 < \sqrt 6 < 2.5 \\ $$
imply
$$-0.9 < \sqrt{6}-\sqrt{2}-\sqrt{3} <-0.6$$
and so $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is not an integer. Therefore, it is irrational.