Two different maximum!

The first part of the problem is the same as one of my problems: x, y and z

From that is not so difficult to get that $B=A$ , and since $A=\dfrac{4}{27}$ ,

$54(A+B)=\boxed{16}$

$\underline{x^2y+y^2z+z^2x\le\dfrac{4}{27}}$ Without Loss of Generality, assume $x\ge y\ge z$

$2(x^2y+y^2z+z^2x)-x(x+z)(2y+z)=2yz(y-x)+xz(z-x)\le 0$

$\therefore x^2y+y^2z+z^2x\le \dfrac{1}{2}x(x+z)(2y+z)\le \dfrac{1}{2}\left(\dfrac{x+x+z+2y+z}{3}\right)^3=\dfrac{4}{27}$

Equality holds iff $x=\dfrac{2}{3},y=\dfrac{1}{3},z=0$

$\underline{x^2y+y^2z+z^2x+xyz\le\dfrac{4}{27}}$ Without Loss of Generality, assume $x\ge z\ge y$

$(x-z)(y-z)\le 0$

$xy+z^2\le xz+yz$

$x^2y+xz^2\le x^2z+xyz$

$x^2y+y^2z+z^2x+xyz\le x^2z+2xyz+y^2z=z\left(x+y\right)^2\le \dfrac{4\left(z+\dfrac{x+y}{2}+\dfrac{x+y}{2}\right)^3}{27}=\dfrac{4}{27}$

Equality holds iff $x=y=z=\dfrac{1}{3}$