Prove $x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$ for $xy+yz+zx=1$ 
Let $x,y,z \geqslant 0$ and $xy+yz+zx=1$. Prove that
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x.$$

What I try:
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$$
$$\Leftrightarrow x(xy+yz+zx)+y^2+z^3- x^2y-y^2z-z^2x \geqslant 0$$
$$\Leftrightarrow \left(x^2z+\frac{z^3}{4}-z^2x \right)+ \left(y^2+\frac{3z^3}{4}+xyz-y^2z\right) \geqslant 0$$
I strongly believe that $\left(y^2+\frac{3z^3}{4}+xyz-y^2z\right) \geqslant 0$, but I have no proof.
 A: As you wrote, from the identity:
$$x(xy+yz+zx)+y^2+z^3-x^2y-y^2z-z^2x=z(\frac{z}{2}-x)^2+(y^2+\frac{3}{4}z^3-y^2z)+xyz$$
It suffices to check $y^2+\frac{3}{4}z^3\ge y^2z$ given $yz\le 1$. We take $3$ cases:
If $z\le 1$, then $y^2\ge y^2z$ already.
If $z\ge 1$ and $y\le \frac{1}{4}$, then $\frac{3}{4}z^3\ge \frac{3}{4}z\ge \frac{1}{16}z\ge y^2z$.
If $z\ge 1$ (so $y\le 1$) and $y\ge\frac{1}{4}$ (so $z\le 4$) then $y^2\ge \frac{1}{4}y^2z$ and $\frac{3}{4}z^3\ge \frac{3}{4}z\ge \frac{3}{4}y^2z$. Adding these gives the required.
This covers all cases, so the inequality is proved. "Equality" holds when $x=\frac{1}{\epsilon}, y=\epsilon, z=0$ and $\epsilon\to 0$. 
A: Alternative proof:
Using $xy + zx \le 1$ and $\sqrt{yz} \le 1$, we have
\begin{align*}
 &x + y^2 + z^3 - x^2 y - y^2z - z^2x \\
 \ge\,& x(xy + zx) + y^2\sqrt{yz} + z^3 - x^2 y - y^2z - z^2x\\
 =\,& x^2z + y^2\sqrt{yz} + z^3 - y^2z - z^2x \\
 =\,& \left(x^2z - z^2x + \frac14 z^3\right) + \frac34 z^3 + y^2\sqrt{yz} - y^2z\\
 =\,& \frac14 z(2x - z)^2 + \frac34 z^3 + y^2\sqrt{yz} - y^2z \\
 \ge\,& \frac14 z(2x - z)^2 + \frac34 z^3 + y^2\cdot \frac{2yz}{y + z} - y^2z \\
 =\,& \frac14 z(2x - z)^2 + \frac{z}{4(y+z)}(4y^3 + 3yz^2  - 4y^2z + 3z^3) \\
 \ge\,& 0 
\end{align*}
where we have used GM-HM to get
$\sqrt{yz} \ge \frac{2yz}{y + z}$ (note: $y + z > 0$),
and AM-GM to get
$4y^3 + 3yz^2  - 4y^2z \ge 2\sqrt{4y^3 \cdot 3yz^2} - 4y^2z = (4\sqrt 3 - 4)y^2z \ge 0$.
We are done.
