TITLE: $x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
QUESTION [40 upvotes]: $x,y,z \geqslant 0$, $x+y^2+z^3=1$, 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.

REPLY [2 votes]: $\color{green}{\textbf{Computer free version of 17.12.20.}}$
$\color{brown}{\textbf{Preliminary notes.}}$
We should prove the inequality$\;f(x,y,z)= x^2y+y^2z+z^2x \le \frac12\;$ under the conditions
$x+y^2+z^3=1,\;$ $x,y,z\in [0,1].$
In accordance with the @RiverLi answer,


$y\le y^2+\frac14,\tag1$
$2-4(x^2y+y^2z+z^2x)\ge 2-x^2-(4x^2+4z)(1-x-z^3)-4z^2x,\tag2$


where $(1)$ follows from the evident $\;\left(y-\frac12\right)^2 \ge0\;$.
At the same time, for $\;t\in[0,1]\;$
$$t^4-t+\frac3{4\sqrt[\large3]4} = \left(t^2-\sqrt[\large3]2t-\frac1{2\sqrt[\large3]2}\right)\left(t^2 +\sqrt[\large3]2 t +\frac3{2\sqrt[\large3]2}\right) $$
$$= \left(t-\frac1{\sqrt[\large3]4}\right)^2\left(t^2+\sqrt[\large3]2 t 
+\frac3{2\sqrt[\large3]2}\right) \ge 0,$$
$$t\le t^4+H,\quad \text{where}\quad H=\frac3{4\sqrt[\large3]4} = \dfrac12\sqrt[\large3]{\dfrac{27}{32}}<\dfrac12,\quad H\approx0.47247.\tag3$$
$\color{brown}{\textbf{The edges of area.}}$
Taking in account $(1)$ and $(3),$ one can get
\begin{align}
&f(x,y,0) = x^2y \le (1-y^2)^2y\,\bigg|_{\large y=\frac1{\sqrt5}} = \dfrac{16}{25\sqrt5}<0.28622,  \\[4pt]
&f(x,0,z) = z^2x = (1-z^3)z^2 \le (1-z^3)z \le H, \\[4pt]
&f(0,y,z) = y^2z = (1-z^3)z \le H.
\end{align}
Thus, the given inequality is correct on the edges of the area.
$\color{brown}{\textbf{The inner stationary points.}}$
Taking in account $(2)$ and $(1)$ in the form of
$\;4z\le 4z^2+1,\;$
it suffices to prove the inequality $\;g(x,z)\ge0,\;$ where
$$g(x,z) = 2-x^2-(4x^2+1+4z^2)(1-x-z^3)-4z^2x,$$
$$g(x,z) = 1+4x^2z^3 + (4z^5+z^3-4z^2) + (4x^3-5x^2+x),\tag4$$
under the conditions $\;\;x,z\in(0,1).$
Stationary points should satisfy the system $\;g'_x = g'_z =0,\;$
or
\begin{cases}
g'_x = 8z^3x + 12 x^2 - 10x + 1 = 0\\[4pt]
g'_z = 12x^2z^2+20 z^4+3z^2-8z = 0,\tag5
\end{cases}
and then
\begin{cases}
12x^2z = 8-3z-20z^3\hspace{163mu} \left[z^{-1}\times(5.2)\right]\\
8z^4x +(8-3z-20z^3) - 10xz + z = 0,\qquad \left[z\times(5.1)\right]\\
\end{cases}
$$x = h_1(z) = \dfrac{4-z-10z^3}{5z-4z^4},\quad x^2 = h_2(z) = \dfrac{8-3z-20z^3}{12z}\tag6.$$
At the same time, from $(4),(5)$ should
\begin{align}
&5g(x,z) = 5g - xg'_x - z g'_z\\[4pt]
& = 5+20x^2z^3 + (20z^5+5z^3-20z^2) + (20x^3-25x^2+5x)\\[4pt]
& - x(8z^3x + 12 x^2 - 10x + 1) - z(12x^2z^2+20 z^4+3z^2-8z)\\[4pt]
& = 5 + 4 x + 8x^3 -15x^2 - 12 z^2 + 2 z^3,
\end{align}
$$5g(x,z) = 5 - 15h_2(z) + 4 h_1(z) (1+2h_2(z)) - 12 z^2 + 2 z^3.\tag7$$
Substitution of $(6)$ to $(7)$ represents the inequality $\;g(x,z) \ge 0\;$ in the form of
$$256 - 568 z + 501 z^2 - 1280 z^3 + 1180 z^4 - 300 z^5 + 1600 z^6 - 624 z^7 - 96 z^8\ge 0.\tag8$$
Since
$$h(z)-1=\dfrac{4z^4-10z^3-6z+4}{5z-4z^4}=\dfrac{2(1-2z)(2+z+2z^2-z^3)}{5z-4z^4},$$
then $\;h(z)>1,\;$ if $\;z\in\left(0,\frac12\right).\;$
Besides, $\;h(z)\;$ has a positive denominator and  decreasing numerator in the interval $\;z\in(0,1],\;$ wherein $\;h\left(\frac7{10}\right) = -\frac{325}{6349}<0.\;$
Therefore, $\;h(z)< 0,\;$ if $\;z\in\left[\frac7{10},1\right].\;$
Thus, all inner stationary points belong to the interval $\;z\in\left[\dfrac12,\dfrac7{10}\right).\;$
And now, substitution $\;z=\dfrac{6+t}{10},\;t\in[-1,1),\;$ transforms $(8)$ to the evident inequality
$$-3 t^8 - 339 t^7 - 6214 t^6 - 13083 t^5 + 1041140 t^4\\ + 12923432 t^3 + 64494426 t^2 - 898904 t + 13278632 \ge 0.$$
Done!