Áp dụng BĐT:
\(xyz\ge\left(x+y-z\right)\left(y+z-x\right)\left(x+z-y\right)\)
\(\Leftrightarrow xyz\ge\left(1-2x\right)\left(1-2y\right)\left(1-2z\right)\)
\(\Leftrightarrow xyz\ge1+4\left(xy+yz+zx\right)-2\left(x+y+z\right)-8xyz\)
\(\Leftrightarrow9xyz\ge4\left(xy+yz+zx\right)-1\)
\(\Rightarrow P=x^2+y^2+z^2+\frac{9}{2}xyz\ge x^2+y^2+z^2+2\left(xy+yz+zx\right)-\frac{1}{2}\)
\(\Leftrightarrow P\ge\left(x+y+z\right)^2-\frac{1}{2}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
Lại có:
\(xy+yz+zx=\left(xy+yz+zx\right)\left(x+y+z\right)\ge3\sqrt[3]{x^2y^2z^2}.3\sqrt[3]{xyz}=9xyz\)
\(\Rightarrow P\le x^2+y^2+z^2+\frac{1}{2}\left(xy+yz+zx\right)\)
\(P\le\left(x+y+z\right)^2-\frac{3}{2}\left(xy+yz+zx\right)\le\left(x+y+z\right)^2=1\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và hoán vị
