Ta có: \(x^3+y^3=3xyz-z^3\)
\(\Leftrightarrow\left(x^3+y^3\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow3\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}}\left(\forall x,y,z\right)\Rightarrow Vt\ge0\left(\forall x,y,z\right)\)
Dấu "=" xảy ra khi: \(\left(x-y\right)^2=\left(y-z\right)^2=\left(z-x\right)^2=0\)
\(\Rightarrow x=y=z=1\)
Từ đó \(P=673\cdot\left(1^{2020}+1^{2020}+1^{2020}\right)+1=2020\)
Vậy P = 2020
