n is a natural unzero number, find x natural so that:
$\displaystyle \sqrt{x-1}<\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}<\sqrt{x}$
I saw that x must be bigger than n, but how I find it?
As seen in veileen's post, there's an abundance of solutions to this inequality.
Therefore, I think that all you can do is simplify this expression a bit by quadrating the whole thing, so that you can get a clearer relationship between n and x.
If you quadrate the whole inequality, I think that this will be the end result:
$\displaystyle x-1<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x$
but I might have been a bit sloppy while I did it, so I can be a little bit off.
Do you mean like something like this:
$\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1$
Where the only natural number for a given n, which makes the inequality true, is x.
I personally don't know how I should go about determining the exact value of x, i.e. x= 'something', when I'm only given this type of inequality, which contains many different solutions.
$\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1$
$\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})$ is not natural ~ "I personally don't know how I should go about determining the exact value of x," - using integer part. (?)
Does it have to be? - No, that is why I suggested use the integer part.
$\displaystyle 3n+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]}\leq 3n+2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})<x$
$\displaystyle x<3n+1+2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})\leq3n+1+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]+1$
$\displaystyle =3n+2+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]$
$\displaystyle [2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]\in \mathbb{N} \Rightarrow x=3n+1+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]$ (because it's a natural number between $\displaystyle 3n+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]}$ and $\displaystyle 3n+2+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})])$
* [] - means integer part