Ta có: $1=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx) $nên $P=x+y+z+\frac{(x+y+z)^{2}-1}{2}=\frac{1}{2}(x+y+z+1)^{2}-1 \geq -1 $Khi $x=-1, y=z=0 $ thì $P=-1$. Vậy $\min P=-1$Ta có: $xy+yz+zx \leq x^{2}+y^{2}+z^{2}=1 $Nên $(x+y+z)^{2}=1+2(xy+yz+zx)\leq 3 $. Do đó $P \leq \sqrt{3}+1 $Khi $x=y=z=\frac{1}{\sqrt{3} } $ thì $P=\sqrt{3} $. Vậy $\max P=\sqrt{3}+1 $
khó
Ta có: $1=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx) $nên $P=x+y+z+\frac{(x+y+z)^{2}-1}{2}=\frac{1}{2}(x+y+z+1)^{2}-1 \geq -1 $Kh
i $x=-1, y=z=0 $ thì $P=-1$. Vậy $\min P=-1$Ta có
: $xy+yz+zx \leq x^{2}+y^{2}+z^{2}=1 $Nên $(x+y+z)^{2}=1+2(xy+yz+zx)\leq 3 $. Do đó $P \leq \sqrt{3}+1 $Khi $x=y=z=\frac{1}{\sqrt{3} } $ thì $P=\sqrt{3} $. Vậy $\max P=\sqrt{3}+1 $