Đặt y = \(\sqrt{x-1}\)\(\ge0\Rightarrow\)\(y^2=x-1\Rightarrow x=y^2+1\)
\(x+\sqrt{x-1}=y^2+1+y=y^2+2.\frac{1}{2}y+\frac{1}{4}+\frac{3}{4}\)
\(=\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\ge\left(\frac{1}{2}\right)^2+\frac{3}{4}=1\)
Dấu "=" xảy ra khi \(y=\sqrt{x-1}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)