$x,y,z \geqslant 0$, $x+y^2+z^3=1$, 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.

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 $p$. 
