<s>[INST] $x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$ 
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove 
  $$x^2y+y^2z+z^2x < \frac12$$

This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.
 [/INST] I don't think the best bound can be solved for analytically.
using Lagrange multipliers , I tried to maximize $x^2y+y^2z+z^2x$ $\quad$ subject to the constraint $x+y^2+z^3=1$
maximize $g=x^2y+y^2z+z^2x+\lambda(x+y^2+z^3-1)$
$\frac {\partial g}{\partial x}=0:2xy+z^2+\lambda=0$
$\frac {\partial g}{\partial y}=0:x^2+2yz+2\lambda y=0$
$\frac {\partial g}{\partial z}=0:y^2+2zx+3\lambda z^2=0$
$\frac {\partial g}{\partial \lambda}=0:x+y^2+z^3-1=0$
$x\frac {\partial g}{\partial x}=0:2x^2y+z^2x+\lambda x=0$
$y\frac {\partial g}{\partial y}=0:x^2y+2y^2z+2\lambda y^2=0$
$z\frac {\partial g}{\partial z}=0:y^2z+2z^2x+3\lambda z^3=0$
$3(x^2y+y^2z+z^2x)=-\lambda (x+2y^2+3z^3)=(2xy+z^2)(x+2y^2+3z^3)$
If the global maximum occur on $ x, y, z \ge 0 $ then we can solve $(2xy+z^2)(x+2y^2+3z^3) \ge 0 $ for $x, y, z \ge 0 $ 
If no local maximum occur at $ x, y,z \ge 0 $ then we have to search for farthest points from the point of global minimum.
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