radicals - Proof that $sqrt6 - sqrt2 - sqrt3$ is irrational.

I want to prove that: $$\sqrt6 - \sqrt2 - \sqrt3$$ is irrational. I have tried using squares, the $p/q$ definition of rationality and the facts that

1)rational$\times$ irrational=irrational (unless rational=0),

2)rational$+$irrational=irrational.

However, I haven't been able to reach some conclusion. Things seem harder than when you have two square roots. Any help would be appreciated!

Answer

Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational.

Then, $(\sqrt{3}-1)(\sqrt{2}-1)=\sqrt{6}-\sqrt{2}-\sqrt{3}+1$ is a rational number, say $r\in\mathbb{Q}$.
That is, $\sqrt{3}-1=\frac{r}{\sqrt{2}-1}=r(\sqrt{2}+1)$.
Thus, $\sqrt{3}-r\sqrt{2}=r+1\in\mathbb{Q}$.

Clearly, $r\neq -1$, whence $\sqrt{3}-r\sqrt{2}\neq 0$.
Now, $\sqrt{3}+r\sqrt{2}=\frac{3-2r^2}{\sqrt{3}-r\sqrt{2}}=\frac{3-2r^2}{r+1}\in\mathbb{Q}$.

What happens if both $\sqrt{3}-r\sqrt{2}$ and $\sqrt{3}+r\sqrt{2}$ are rational numbers?

This line of reasoning shows that $a\sqrt{pq}+b\sqrt{p}+c\sqrt{q}$ is irrational if $a,b,c\in\mathbb{Q}$ with $a\neq0$ and $p,q\in\mathbb{N}\setminus\{1\}$ are such that $p$ and $q$ are distinct and square-free.