\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{y+x-3}{z}=\frac{y+z+1+x+z+2+y+x-3}{x+y+z}=\frac{2x+2y+2z}{x+y+z}=2\)

=>\(\begin{cases}y+z+1=2x\\ x+z+2=2y\\ x+y-3=2z\end{cases}\Rightarrow\begin{cases}y+z=2x-1\\ x+z=2y-2\\ x+y=2z+3\end{cases}\)

Ta có: \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{y+x-3}{z}=\frac{1}{x+y+z}\)

=>\(\frac{1}{x+y+z}=2\)

=>\(x+y+z=\frac12\)

Ta có: \(x+y+z=\frac12\)

=>\(2x-1+x=\frac12\)

=>\(3x=\frac32\)

=>\(x=\frac12\)

Ta có: \(x+y+z=\frac12\)

=>\(2y-2+y=\frac12\)

=>\(3y=2+\frac12=\frac52\)

=>\(y=\frac56\)

Ta có: \(x+y+z=\frac12\)

=>\(2z+3+z=\frac12\)

=>\(3z=\frac12-3=-\frac52\)

=>\(z=-\frac56\)

\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)

\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\) 

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)