Ta có: (x−y)2+(y−z)2+(z−x)2=2(x2+y2+z2)−2(xy+yz+zx)≥0⇒xy+yz+zx≤1(x+y+z)2=x2+y2+x2+2(xy+yz+zx)≤3(x2+y2+z2)=3⇒x+y+z≤√3
Từ đó suy ra: x+y+z+xy+yz+zx≤1+√3
Vậy MaxP=1+√3 khi x=y=z=√33
Mặt khác ta lại có (x+y+z)2=1+2(xy+yz+zx)⇒xy+yz+zx=(x+y+z)2−12
nên P=x+y+z+(x+y+z)2−12=12(x+y+z+1)2−1≥−1
Vậy
MinP=−1 khi
x=−1,y=z=0