ta có \(\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}\)
\(=\sqrt{\left(\sqrt{x-2}+1\right)^2}+\sqrt{\left(\sqrt{x-2}-1\right)^2}\)
\(=\left|\sqrt{x-2}+1\right|+\left|\sqrt{x-2}-1\right|\)
Vì \(x\ge2\Rightarrow\sqrt{x-2}\ge0\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2}+1\ge1\\\sqrt{x-2}-1\ge-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|\sqrt{x-2}+1\right|\ge1\\\left|\sqrt{x-2}-1\right|\ge1\end{matrix}\right.\)
\(\Leftrightarrow\left|\sqrt{x-2}+1\right|+\left|\sqrt{x-2}-1\right|\ge2\)
Hay A\(\ge2\) Dấu = xảy ra khi x=2
=> đpcm
