\(\Rightarrow x^2+y^2=9-z^2\ge0\Leftrightarrow-3\le z\le3\)
\(\Rightarrow P=\dfrac{x+z-1}{y+4}\ge\dfrac{x-3-1}{y+4}=\dfrac{x-4}{y+4}=\dfrac{3+z-y-4}{y+4}=\dfrac{-y+z-1}{y+4}\ge\dfrac{-y-3-1}{y+4}=\dfrac{-\left(y+4\right)}{y+4}=-1\)
dấu"=" xảy ra \(\Leftrightarrow x=y=0;z=-3\)