$x,y,z \geqslant 0$, $x+y^2+z^3=1$, 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.
 A: This answer is incomplete.
Let $A=x+y^2$, $B=y^2+z^3$ and $C=z^3+x$.
Claim: $x^2y \leq \dfrac{A}{2}x^{3/2}$, 
$y^2z \leq \dfrac{B^{2/3}}{4^{2/3}}y^{4/3}$, $z^2x \leq \dfrac{C^{4/3}}{4^{2/3}}x^{1/3}$.
Assuming the claim, we note that $A+B+C=2$ and we need to prove that the given sum is less than $\dfrac{A+B+C}{4}$.
A: 
The method to be employed here is exactly the same as in a previous answer:

Olympiad inequality $\sum_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

Let $u = x$ , $v = y^2$ , $w = z^3$ , then $u,v,w \ge 0$ , $u+v+w=1$ and the inequality to be established:
$$
x^2y+y^2z+z^2x < \frac12 \quad \Longrightarrow \quad f(u,v,w) = u^2v^{1/2}+v\,w^{1/3}+w^{2/3}u < \frac12
$$
The maximum of this function inside the equilateral triangle must shown to be less than $1/2$.
There is no symmetry argument, because the latter is effectively destroyed by the "weird" condition $x+y^2+z^3=1$ , as it is called.
Another proof without words is attempted
by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:


nivo := min + sqrt(g/grens)*(max-min); { sqrt = square root ; grens = 25 ; g = 0..grens }

The darkness of the isolines is proportional to the (positive) function values; they are almost black near the maximum and almost white near the minimum values.
Maximum and minimum values of the function are observed to be:


 4.58251457205350E-0003 < f < 4.85621276951755E-0001 < 1/2

The little $\color{blue}{\mbox{blue}}$ spot is where $\left|f(u,v,w) - \mbox{max}\right|< 0.0002$ .
This maximum is close to values found by other people here, but not quite.
Perhaps it's interesting to know the position of the maximum as well:


(x,y,z) = ( 5.58304528246164E-0001 , 4.97693736095187E-0001 , 5.78892473099889E-0001 )

I have no idea how to convert these numerical values into something more analytical.

EDIT. What you see is what you get :-) Without doubt. Some decent error analysis reveals that the value of the maximum to be trusted in this answer is : $\;0.48562 \pm 0.00003\;$ , quite in agreement with values given elsewhere (e.g. in the comment by Steve Kass).
A: 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.
A: Let $x=0$.
Then $y^2+z^3=1 \Rightarrow z^3=1-y^2$
Seek to maximise $p=y^2z=y^2(1-y^2)^{\frac 13}$
WLOG let $u=y^2$ so that $p=u(1-u)^{\frac 13}$
$\frac {dp}{du}=(1-u)^{\frac 13}-u{\frac 13}(1-u)^{-\frac 23}$
$\frac {dp}{du}=\frac 13(3-4u)(1-u)^{-\frac 23}$
Maximum is at $u=\frac 34$
Thus maximum is $p=\frac 34(\frac 14)^{\frac 13}$
$p \le \frac 34(\frac 14)^{\frac 13}$
$p^3 \le \frac {27}{64}\frac 14$
$p^3 \le \frac {27}{256} < \frac {32}{256}$
$p^3 < \frac {1}{8}$
$p < \frac {1}{2}$
We then have to demonstrate that any increase in $x$ creates corresponding decreases in $y^2$ and $z^2$ such that the other two terms in the expression $x^2y+y^2z+z^2x$ increase more slowly than the decrease in $p$. 
A: More idea than answer, but it might be of help.
If we replace $x$ by $x^2$ and $z$ by $z^{2/3}$, the constraint becomes $x^2+y^2+z^2=1$ and the inequality becomes
$$x^4y+y^2z^{2/3}+z^{4/3}x^2\lt{1\over2}$$
This looks worse, but the advantage of constraining $(x,y,z)$ to the unit sphere is that we can switch over to spherical coordinates, writing $x=\sin\theta\cos\phi$, $y=\sin\theta\sin\phi$, and $z=\cos\theta$, and make use of some trig identities.  Ultimately, with a final change of variable $u=(\cos\theta)^{2/3}$ and $v=-\cos2\phi$, the inequality to prove (if I've done the algebra correctly) is
$$(1-u^2)^{5/2}\left({(1-v)(1+v)^4\over8}\right)^{1/2}+(u-u^4)(1+v)+(u^2-u^5)(1-v)\lt1\quad\text{for }0\le u,v\le1$$
This "reduces" the problem to the unpleasant chore of finding (or bounding) a  function of two variables over the unit square.  I'll end by noting that if we focus on the easy portion of the function, $f(u,v)=(u-u^4)(1+v)+(u^2-u^5)(1-v)$, we find that $f$ is maximized when $v=1$ and $u=1/\sqrt[3]4$, and
$$f\left({1\over\sqrt[3]4},1\right)={3\over2\sqrt[3]4}\approx0.94494$$
so the true minimum of the full function is not much less than $1$. (Just a note: I multiplied both sides of the original inequality by $2$ to clear out some denominators.  Dividing by $2$ changes $0.94494$ to $0.47247$, which is, as it should be, less than the numerical result $0.48562209920309984177$ reported by Steve Kass in comments below the OP.)
