Ta có: $(x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2)-2(xy+yz+zx)\geq 0\Rightarrow xy+yz+zx\leq 1$$(x+y+z)^2=x^2+y^2+x^2+2(xy+yz+zx)\leq 3(x^2+y^2+z^2)=3\Rightarrow x+y+z\leq \sqrt{3}$
Từ đó suy ra: $x+y+z+xy+yz+zx\leq 1+\sqrt{3}$
Vậy $Max P=1+\sqrt{3}$ khi $x=y=z= \frac{\sqrt{3}}{3}$
Mặt khác ta lại có $(x+y+z)^2=1+2(xy+yz+zx)\Rightarrow xy+yz+zx=\frac{(x+y+z)^2-1}{2}$
nên $P=x+y+z+\frac{(x+y+z)^2-1}{2}=\frac{1}{2}(x+y+z+1)^2-1\geq -1$
Vậy $MinP=-1$ khi $x=-1,y=z=0$