\(P=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

Condition : \(x\ge1\)

a) We have \(x+2\sqrt{x-1}=\left(x-1\right)+2+1\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\)

So P = \(\sqrt{\left(1+\sqrt{x-1}\right)^2}+\sqrt{\left(1-\sqrt{x-1}\right)^2}\)

\(\Rightarrow\left|1+\sqrt{x-1}\right|+\left|1-\sqrt{x-1}\right|\)

+ With \(1-\sqrt{x-1}\ge0\Rightarrow x\le2:\)

\(P=1+\sqrt{x-1}+1-\sqrt{x-1}=2\)

+ With \(1-\sqrt{x-1}< 0\Rightarrow x>2:\)

\(P=1+\sqrt{x-1}+\sqrt{x-1}-1=2\sqrt{x-1}\)

Sum up : \(\left[{}\begin{matrix}1\le x\le2\Rightarrow P=2\\x>2\Rightarrow P=2\sqrt{x-1}\end{matrix}\right.\)

b) We have inequality \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

So : \(P=\left|1+\sqrt{x-1}\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{\left(1-\sqrt{x-1}\right)+\left(1-\sqrt{x-1}\right)}\right|\)

\(\Rightarrow P\ge2\Leftrightarrow1\le x\le2\)

So Pmin = 2 when \(1\le x\le2\).