simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$ 
Simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$

The problem is from a not so well renowned book for calculus in India - Concepts of Functions & Calculus - Vikas Rahi, ISBN 9780070080805. The answer for this question is given as
$$\lvert \sqrt{x-1} -1  \rvert + \lvert \sqrt{x-1}+1 \rvert$$
My efforts weren't great at all, I tried rationalizing them but the denominator becomes very similar to the question, which is totally not helpful for simplification.
 A: Write $t =\sqrt{x-1}$ then your expression is $$\sqrt{t^2+1+2t} +\sqrt{t^2+1-2t} =|t+1|+|t-1| $$
Notice that $t\geq 0$. So if $t\geq 1$ we get $2t$ and if $t<1$ we get $2$.
A: First, note that
\begin{align}
(1 \pm \sqrt{x-1})^2
&= 1 \pm 2 \sqrt{x-1} + \sqrt{x-1}^2 \\
&= 1 \pm 2 \sqrt{x-1} + x-1 & x \geq 1 \\
&= x \pm 2 \sqrt{x-1}
\end{align}
Thus, since $\sqrt{u^2} = |u|$,
\begin{align}
f(x)
&= \sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}} \\
&= \sqrt{(1-\sqrt{x-1})^2}+\sqrt{(1+\sqrt{x-1})^2} & x \geq 1 \\
&= |1-\sqrt{x-1}|+|1+\sqrt{x-1}| \\
&= \begin{cases}
(1-\sqrt{x-1})+(1+\sqrt{x-1}) & 1 \leq x < 2 \\
-(1-\sqrt{x-1})+(1+\sqrt{x-1}) & 2 \leq x
\end{cases} \\
&= \begin{cases}
2 & 1 \leq x < 2 \\
2 \sqrt{x-1} & 2 \leq x
\end{cases}
\end{align}
A: Just write
\begin{eqnarray*} \sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}
& = & \sqrt{x -1 - 2\sqrt{x -1} +1} +\sqrt{x -1 + 2\sqrt{x -1} +1} \\
& = & \sqrt{(\sqrt{x-1}-1)^2} + \sqrt{(\sqrt{x-1}+1)^2} \\
& = & \begin{cases} 2 & 1 \leq x\leq 2 \\ 2\sqrt{x-1} & x>2 \end{cases} 
\end{eqnarray*}
A: $$\begin{align}{\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}&=T(x)>0&\end{align}$$
$$T^2(x)=2x+2 \sqrt{x^2-4x+4}$$
$$T^2(x)=2x+2|x-2|$$
$$\begin{align}\color {gold}{\boxed {\color{black}{T(x)=\sqrt{2x+2|x-2|}}}}\end{align}$$


*

*If $x≥2$, then $\begin{align}T(x)&=\sqrt{2x+2x-4}\\
&=2\sqrt{x-1}\end{align}$


*If $x<2$, then $\begin{align}T(x)=\sqrt{2x+4-2x}=2.\end{align}$
A: "Simplify" can easily depend on the beholder's eyes, but one idea (and hint) is:
$$\sqrt{a-b}+\sqrt{a+b}=\frac{2b}{\sqrt{a+b}-\sqrt{a-b}}$$
I  don't really think we get a huge simplifaction here but...perhaps something like this is what is meant in that book.
