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?

Let's use Mathematica for an overview plot:
ReImPlot[Sqrt[x + 2 Sqrt[x] - 1] + Sqrt[x - 2 Sqrt[x] - 1],
         {x, -1, 7}, PlotPoints -> 10^3, 
         GridLines -> {{0, 3 - 2 Sqrt[2], 3 + 2 Sqrt[2]},
                       {0, 2 Sqrt[Sqrt[2] - 1], 2, 2 Sqrt[1 + Sqrt[2]]}}, 
         PlotTheme -> "Scientific"]



*

*For $x<0$ the function is purely real.

*For $0\le x\le 3-2\sqrt{2}$ the function is purely imaginary.

*For $3-2\sqrt{2}<x<3+2\sqrt{2}$ the function has nonvanishing real and imaginary parts.

*For $x\ge3+2\sqrt{2}$ the function is purely real.

