\(P=a^3+b^3+c^3+a^2\left(b+c\right)+b^2\left(a+c\right)+c^2\left(a+b\right)\)
\(=a^3+b^3+c^3+a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\)
\(=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
Đạt được khi \(a=b=c=\frac{1}{3}\)