$P^2=\frac94x^2+\frac{16}9y^2+\frac{25}{36}z^2+4xy+\frac{20}9yz+\frac52zx$$=(\frac94x^2+\frac14z^2)+(\frac{16}9y^2+\frac49z^2)+4xy+\frac{20}9yz+\frac52zx$$\ge\frac32xy+\frac{16}9yz+4xy+\frac{20}9yz+\frac52zx$$=4(xy+yz+zx)=4$Dấu bằng xảy ra $\begin{cases}z=2y=3x \\ xy+yz+zx=1 \end{cases}$
$P^2=\frac94x^2+\frac{16}9y^2+\frac{25}{36}z^2+4xy+\frac{20}9yz+\frac52zx$$=(\frac94x^2+\frac14z^2)+(\frac{16}9y^2+\frac49z^2)+4xy+\frac{20}9yz+\frac52zx$$\ge\frac32xy+\frac{16}9yz+4xy+\frac{20}9yz+\frac52zx$$=4(xy+yz+zx)=4$Dấu bằng xảy ra $\begin{cases}z=2y=6x \\ xy+yz+zx=1 \end{cases}\Leftrightarrow \begin{cases}x=\frac1{3\sqrt3} \\ y=\frac1{\sqrt3}\\z=\frac2{\sqrt3} \end{cases}$
$P^2=\frac94x^2+\frac{16}9y^2+\frac{25}{36}z^2+4xy+\frac{20}9yz+\frac52zx$$=(\frac94x^2+\frac14z^2)+(\frac{16}9y^2+\frac49z^2)+4xy+\frac{20}9yz+\frac52zx$$\ge\frac32xy+\frac{16}9yz+4xy+\frac{20}9yz+\frac52zx$$=4(xy+yz+zx)=4$Dấu bằng xảy ra $\begin{cases}z=2y=
3x \\ xy+yz+zx=1 \end{cases}$