Normal
\[\huge \begin{cases} 5-z=y+x\\ 8=z(y+x)+yx\le z(y+x) +\frac{(y+x)^2}{4}\end{cases} \]\[\huge \rightarrow \frac{(5-z)^2}{4}+z(5-z) \ge 8 \]\[\huge \rightarrow \frac{(3z-5)^2}{12} \le \frac{1}{3} \]\[\huge \rightarrow 1 \le z \le \frac{7}{3} \]Do tính đối xứng nên ta có \[\huge \rightarrow 1 \le x,y,z \le \frac{7}{3} \]
\[\huge \begin{cases} 5-z=y+x\\ 8=z(y+x)+yx\le z(y+x) +\frac{(y+x)^2}{4}\end{cases} \]
\[\huge \rightarrow \frac{(5-z)^2}{4}+z(5-z) \ge 8 \]
\[\huge \rightarrow \frac{(3z-5)^2}{12} \le \frac{1}{3} \]
\[\huge \rightarrow 1 \le z \le \frac{7}{3} \]
Do tính đối xứng nên ta có
\[\huge \rightarrow 1 \le x,y,z \le \frac{7}{3} \]
