ĐKXđ: \(y\ge z\ge x\ge0\)
\(\sqrt{x}+\sqrt{y-z}+\sqrt{z-x}=\dfrac{1}{2}\left(y+3\right)\)
\(\Leftrightarrow y+3-2\sqrt{x}-2\sqrt{y-z}-2\sqrt{z-x}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-z-2\sqrt{y-z}+1\right)+\left(z-x-2\sqrt{z-x}+1\right)=0\)\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-z}-1\right)^2+\left(\sqrt{z-x}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-z}=1\\\sqrt{z-x}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=2\end{matrix}\right.\)
Áp dụng bất đẳng thức AM-GM:
\(\sqrt{x}\le\dfrac{x+1}{2}\)
\(\sqrt{y-z}\le\dfrac{y-z+1}{2}\)
\(\sqrt{z-x}\le\dfrac{z-x+1}{2}\)
Cộng theo vế:
\(\sqrt{x}+\sqrt{y-z}+\sqrt{z-x}\le\dfrac{1}{2}\left(x+1+y-z+1+z-x+1\right)=\dfrac{1}{2}\left(y+3\right)\)
\("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y-z=1\\z-x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\z=2\\y=3\end{matrix}\right.\)
