\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=2\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=2\)
Đặt \(\left\{{}\begin{matrix}x+y+z=a\\x^2+y^2+z^2-xy-yz-zx=b\end{matrix}\right.\) \(\Rightarrow ab=2\Rightarrow b=\frac{2}{a}\)
Mặt khác \(2b=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2>0\) (dấu = ko xảy ra)
\(\Rightarrow a>0\)
Vậy \(P=\frac{a^2}{2}+4b=\frac{a^2}{2}+\frac{8}{a}=\frac{a^2}{2}+\frac{4}{a}+\frac{4}{a}\ge3\sqrt[3]{\frac{16a^2}{2a^2}}=6\)
\(\Rightarrow P_{min}=6\) khi \(a=2\) hay \(x+y+z=2\)
