Điều kiện xác định : \(x\ge0\),\(y\ge1\),\(z\ge2\)
\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Mà \(\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2\ge0\)
Đẳng thức xảy ra khi \(\left(\sqrt{x}-1\right)^2=\left(\sqrt{y-1}-1\right)^2=\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)
