ĐKXĐ: ...
\(2\sqrt{y-1}+2\sqrt{x}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\\\sqrt{z-2}-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)
Giải theo BĐT:
\(VT=1.\sqrt{y-1}+1.\sqrt{x}+1.\sqrt{z-2}\)
\(\Rightarrow VT\le\dfrac{1}{2}\left(1+y-1\right)+\dfrac{1}{2}\left(1+x\right)+\dfrac{1}{2}\left(1+z-2\right)\)
\(\Rightarrow VT\le\dfrac{1}{2}\left(x+y+z\right)\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}1=\sqrt{y-1}\\1=\sqrt{x}\\1=\sqrt{z-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)
