Ta có:
\(\left(\sqrt{8-x}+\sqrt{8-y}+\sqrt{8-z}\right)^2\le3\left(24-x-y-z\right)\)
\(\le3\left(24-\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}\right)=36\)
\(\Rightarrow\sqrt{8-x}+\sqrt{8-y}+\sqrt{8-z}^2\le6\)
Dấu = xảy ra khi \(x=y=z=2\)