Is there any way to prove that $\sqrt {n-1} + \sqrt n + \sqrt {n+1}$ is irrational? Before this is marked as a duplicate I just want to say that I've already read a similar thread, where the original poster asked how they would prove that $\sqrt 2 + \sqrt 5 + \sqrt 7$ is an irrational number. I've read the answers to that thread and I couldn't really &quot;apply&quot;/use them in my situation.

Exercise added for reference. Post was edited because it was referring to a particular case (where n was assigned a value such as 6).
I'm trying to reopen the topic because I am more curious as to how you'd solve this type of exercise in the general form (like in the image).
If  $$r-\sqrt6=\sqrt5+\sqrt7,$$ where $r$ is a rational number, we obtain:
$$r^2+6-2r\sqrt6=12+2\sqrt{35},$$ which gives $$(r^2-6)^2=24r^2+8r\sqrt{210}+140$$ and since $r=0$ is impossible, we obtain a contradiction:
$$\sqrt{210}=\frac{r^4-36r^2-104}{8r}\in\mathbb Q.$$
Just muck.  Isolate radicals one by one to side; square and keep going to eliminate radicals one by one.
If $\sqrt 5 + \sqrt 6 + \sqrt 7 = r\in \mathbb Q$ then
$\sqrt 6 + \sqrt 7 = r- \sqrt 5$ then square both side
$6 + 7 + 2\sqrt{6*7} = r^2 - 2r \sqrt 5 + 5$.
$2\sqrt {6*7} + 2r \sqrt 5= r^2+5-6 -7$ then square both sides
$2^2*6*7 + 2^2r^2*5 + 2*2*2r\sqrt{5*6*7} = (r^2 + 5 -6-7)^2$
Thus $\sqrt {5*6*7} = \frac {(r^2 + 5 -6-7)^2-2^2*6*7-2^2r^2*5}{2*2*2r} \in \mathbb Q$.
And as $5*6*7$ is not a perfect square that is a contradiction.