ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ge0\\y-3\ge0\\z-1\ge0\\\frac{1}{2}\left(x+y+z\right)\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-1\\y\ge3\\z\ge1\\x+y+z\ge0\end{matrix}\right.\)
Ta có : \(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\frac{1}{2}\left(x+y+z\right)\)
=> \(2\sqrt{x+1}+2\sqrt{y-3}+2\sqrt{z-1}=x+y+z\)
=> \(x-2\sqrt{x+1}+y-2\sqrt{y-3}+z-2\sqrt{z-1}=0\)
=> \(x+1-2\sqrt{x+1}+1+y-3-2\sqrt{y-3}+1+z-1-2\sqrt{z-1}+1=0\)
=> \(\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)
Ta thấy : \(\left\{{}\begin{matrix}\left(\sqrt{x+1}-1\right)^2\ge0\\\left(\sqrt{y-3}-1\right)^2\ge0\\\left(\sqrt{z-1}-1\right)^2\ge0\end{matrix}\right.\)
=> \(\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2\ge0\)
- Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x+1}-1=0\\\sqrt{y-3}-1=0\\\sqrt{z-1}-1=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+1=1\\y-3=1\\z-1=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\) ( TM )
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