<s>[INST] Find all real numbers $x$ such that $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number I want to find all values of $x\in \mathbb R$ such that the value of $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number.
I solved it as follows:
$x+2\sqrt{x}-1\ge 0$
$(\sqrt{x}+1)^2-2\ge 0$
$(\sqrt{x}+1)^2\ge 2$
$\sqrt{x}+1\ge 2$ or $\sqrt{x}+1\le -\sqrt{2}$
The second can't hold, from the first $x\ge 3-2\sqrt{2}$.
Similarly $(\sqrt{x}-1)^2\ge 2$, hence $x\ge 3+2\sqrt{2}$.
I find my solution to be very ugly. Is my solution correct and is there a neater approach?
 [/INST] The requirement $x\ge0$ is obvious. The expression is real iff $(\sqrt{x}\pm1)^2\ge2$ for both choices of $\pm$, i.e. neither $\sqrt{x}\pm1$ is in $(-\sqrt{2},\,\sqrt{2})$, i.e .$\sqrt{x}\notin[0,\,1+\sqrt{2})\cup[0,\,\sqrt{2}-1)=[0,\,1+\sqrt{2})$, since $\sqrt{x}\ge0$. The solution set is $[3+2\sqrt{2},\,\infty)$, as in @alex.jordan's answer.
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