\(4\left(x^2-x+1\right)\le8x\left(y+z\right)-3x\left(y+z\right)^2\)
\(\Leftrightarrow4\left(x+\frac{1}{x}\right)-4\le8\left(y+z\right)-3\left(y+z\right)^2\)
\(\Rightarrow8\left(y+z\right)-3\left(y+z\right)^2\ge4.2\sqrt{\frac{x.1}{x}}-4=4\)
\(\Leftrightarrow3\left(y+z\right)^2-8\left(y+z\right)+4\le0\)
\(\Leftrightarrow\frac{2}{3}\le y+z\le2\Rightarrow yz\le1\)
\(P=\frac{y^2+3xy\left(x+1\right)}{x^2.yz}+\frac{16}{\left(y+1\right)^3}-10\sqrt{\frac{3y}{x^3+1+1}}\)
\(P\ge\frac{y^2+3xy\left(x+1\right)}{x^2}+\frac{16}{\left(y+1\right)^3}-10\sqrt{\frac{y}{x}}\)
\(P\ge\left(\frac{y}{x}\right)^2+3\left(\frac{y}{x}\right)-10\sqrt{\frac{y}{x}}+3y+\frac{16}{\left(y+1\right)^3}\)
\(P\ge\left(\frac{y}{x}\right)^2+3\left(\frac{y}{x}\right)-10\sqrt{\frac{y}{x}}+6+\left(y+1\right)+\left(y+1\right)+\left(y+1\right)+\frac{16}{\left(y+1\right)^3}-9\)
\(P\ge\left(\sqrt{\frac{y}{x}}-1\right)^2\left(\frac{y}{x}+2\sqrt{\frac{y}{x}}+6\right)+4\sqrt[4]{\frac{16\left(y+1\right)^3}{\left(y+1\right)^3}}-9=-1\)
\(P_{min}=-1\) khi \(x=y=z=1\)
