Ta có:
P=x2+y2+z2+xy+yz+zx
\(\Rightarrow\) 2P= 2x2+2y2+2z2+2xy+2yz+2xz
= (x+y+z)2+x2+y2+z2
= 9+x2+y2+z2
Ta có x2+y2+z2\(\geq\) xy+yz+zx
\(\Leftrightarrow\) 3(x2+y2+z2)\(\geq\) x2+y2+z2+2xy+2yz+2zx
\(\Leftrightarrow\) 3(x2+y2+z2)\(\geq\) (x+y+z)2
\(\Leftrightarrow\) x2+y2+z2\(\geq\) \(\dfrac{\left(x+y+z\right)^2}{3}\) (1)
Từ đó suy ra: 9 + x2+y2+z2\(\geq\) 9+\(\dfrac{\left(x+y+z\right)^2}{3}\) = 9+\(\dfrac{9}{3}\)=9+3=12
\(\Rightarrow\) 2P\(\geq\)12
\(\Rightarrow\) P\(\geq\)6
Dấu = xảy ra khi x=y=z=1
Vậy MinP = 6 khi x=y=z=1
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