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

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    $\begingroup$ When $x=0.534,y=0.5$ and $z=0.6$ (exact decimal values) one has $x^2y+y^2z+z^2x=0.484818$, so 0.5 is close to the best constant. $\endgroup$ May 21, 2016 at 18:19
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    $\begingroup$ Mathematica gives the maximum value of $x^2y+y^2z+z^2x < \frac12$ as constrained to be $0.48562209920309984177$ for $x=0.55811253$, $y=0.49841201$, $z=0.57837131$. $\endgroup$
    – Steve Kass
    May 26, 2016 at 3:23
  • $\begingroup$ I think, this method will be interesting too. $\endgroup$ Apr 6, 2017 at 11:25
  • $\begingroup$ A nice solution is also given by anhduy98@AoPS here: artofproblemsolving.com/community/c6h1238490 $\endgroup$
    – River Li
    Nov 6, 2020 at 2:47

8 Answers 8

18
+25
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The standard way of solving the problem on a conditional extremum is the method of Lagrange multipliers, which reduces it to a system of equations.

The greatest value of function $$f(x,y,z,\lambda) = x^2y+y^2z+z^2x+\lambda(x+y^2+z^3-1)$$ on the interval $$x,y,z\in[0,1]$$ is reached or at its edges, or in the inner stationary point.

$\color{brown}{\textbf{Inner stationary points.}}$

The inner stationary points has zero partial derivatives $$\begin{cases} f'_\lambda = x + y^2 + z^3 - 1 = 0\\ f'_x = z^2 + 2xy + \lambda = 0\\ f'_y = x^2 + 2yz + 2\lambda y = 0\\ f'_z = y^2 + 2zx + 3\lambda z^2 = 0. \end{cases}$$ After the excluding of parameter $\lambda$ get the system $$\begin{cases} x + y^2 + z^3 - 1 = 0\\ x^2 + 2yz = 2y(z^2 + 2xy)\\ y^2 + 2zx = 3z^2(z^2 + 2xy), \end{cases}$$ or $$\begin{cases} x + y^2 + z^3 - 1 = 0\\ (1-3y^2-z^3)^2-4y^4+2yz(1-z)=0\\ 2z(1-3yz)(1-y^2-z^3)+y^2-3z^4=0. \end{cases}$$

Using of Groebner basis allows to get the positive solutions $$ \genfrac{[}{.}{0}{0}{x\approx 0.16367,\quad y\approx 0.761982,\quad z\approx 0.634724,\quad f\approx 0.454882} {x\approx 0.558113,\quad y\approx 0.498412,\quad z\approx 0.578371,\quad f\approx 0.485622}. $$ Alternative way is shown below.

$\color{brown}{\textbf{The edges.}}$

The edges of the field are achieved when $x = 0$, $y = 0$ or $z = 0$.

Substitution $x = 0$ in the expressions for the partial derivatives leads to the system $$ \begin{cases} x=0\\ y^2+z^3 = 1\\ 2yz+2\lambda y = 0\\ y^2+3\lambda z^2 = 0 \end{cases} $$ with solutions $$ \genfrac{[}{.}{0}{} {x=0,\quad y=\sqrt{0.75},\quad z=\sqrt[3]{0.25}\approx 0.629991,\quad f\approx 0.47247} {x=0,\quad y=0,\quad z=1,\quad f=0}$$

Substitution $y = 0$ in the expressions for the partial derivative leads to the system $$ \begin{cases} y=0\\ x+z^3=1\\ z^2+\lambda = 0\\ 2zx+3\lambda z^2 = 0 \end{cases} $$ with solution $$x=0.6,\quad y=0,\quad z=\sqrt[3]{0.4}\approx 0.736806,\quad f\approx 0.32573.$$

Substitution $z = 0$ in the expressions for the partial derivative leads to the system $$ \begin{cases} z=0\\ x+y^2=1\\ 2xy+\lambda = 0\\ x^2+2\lambda y = 0 \end{cases} $$ with solutions $$ \genfrac{[}{.}{0}{} {x=0.8,\quad y=\sqrt{0.2}\approx 0.447214,\quad z=0,\quad f\approx 0.286217} {x=0,\quad y=0,\quad z=1,\quad f=0} $$

The values of function at the vertices of the area (unit parallelepiped) equal to zero.

So the greatest value approximately equals to $0.485622$. Given accuracy of calculations provides the inequality $$\boxed{x^2y+y^2z+z^2x<1/2.}$$

$\color{brown}{\textbf{System resolving, alternative way.}}$

Set condition allows us to reduce the problem to finding unconditional extremes of $$f(y,z)=(1-y^2-z^3)^2y+y^2z+z^2(1-y^2-z^3).$$ Necessary optimality conditions in the field have the form: $$\begin{cases} f'_y = (1-y^2-z^3)^2-4y^2(1-y^2-z^3)+2yz-2yz^2 = 0\\ f'_z = -6yz^2(1-y^2-z^3)+y^2+2z(1-y^2-z^3)-3z^4 = 0, \end{cases}$$ or $$\begin{cases} 5y^4+y^2(6z^3-6)+y(2z-2z^2)+(z^3-1)^2 = 0\\ 6y^3z^2+y^2(1-2z)+6y(z^3-1)z^2-5z^4+2z = 0. \end{cases}$$

If to consider the coefficient of the highest power of $y$ as denominator in equation, and the remaining coefficients - numerators, we get the above equations. Then we can subtract the second equation factor $y$ from the first and repeat subtraction with factor $1$, obtaining the system

$$\begin{cases} C_{2,2}(z)y^2 + C_{2,1}(z)y + C_{2,0}(z) = 0\\ 6y^3z^2+y^2(1-2z)+6y(z^3-1)z^2-5z^4+2z = 0, \end{cases}$$ where $$C_{2,2}(z) = 36z^7-36z^4+20z^2-20z+5,$$ $$C_{2,1}(z) = 18z^6+102z^5-30z^2,$$ $$C_{2,0}(z) = 36z^{10}-72z^7+50z^5+11z^4-20z^2+10z.$$

Thus, the order of the first equation in $y$ reduced from fourth to second. Likewise, lowering the order of the second equation by a first equation, we obtain

$$\begin{cases} C_{2,2}(z)y^2 + C_{2,1}(z)y + C_{2,0}(z) = 0\\ D_{2,2}(z)y^2 + D_{2,1}(z)y + D_{2,0}(z) = 0,\\ \end{cases}\qquad(1)$$ where $$D_{2,2}(z) = 180z^8+576z^7-72z^5-144z^4+40z^3-60z^2+30z-5,$$ $$D_{2,1}(z) = 180z^7-30z^6-30z^5-60z^3+30z^2,$$ $$D_{2,0}(z) = 180z^{11}-252z^8+100z^6-28z^5+25z^4-40z^3+40z^2-10z.$$

The system $(1)$ is a linear in the unknowns $y^2$ and $y$, so $$y^2=\dfrac{\Delta_2(z)}{\Delta_0(z)},\quad y=\dfrac{\Delta_1(z)}{\Delta_0(z)},\qquad(2)$$ where $$\Delta_0(z) = C_{2,2}(z)D_{2,1}(z) - C_{2,1}(z)D_{2,2}(z),$$ $$\Delta_2(z) = -C_{2,0}(z)D_{2,1}(z) + C_{2,1}(z)D_{2,0}(z),$$ $$\Delta_1(z) = -C_{2,2}(z)D_{2,0}(z) + C_{2,0}(z)D_{2,2}(z),$$ or $$\Delta_0(z) = 90z^{10}-828z^9-1662z^8-144z^7+396z^6+1028z^5-200z^4+180z^3-220z^2+10z,$$ $$\Delta_2(z) = -90z^{13}+540z^{12}+30z^{11}+234z^{10}-864z^9-290z^8+256z^7+74z^6+220z^5-160z^4+100z^3-50z^2,$$ $$\Delta_1(z) = 576z^{13}-1296z^{10}+90z^9+815z^8+444z^7-5z^6-580z^5+126z^4+10z^3+80z^2-30z-5.$$

From $(2)$ for $\Delta_0(z)\not=0$ should be $$\Delta_1^2(z) - \Delta_2(z)\Delta_0(z) = 0,$$ or $$331776z^{26}-1484892z^{23}-19440z^{22}+1233720z^{21}+3079404z^{20}+195732z^{19}-3189924z^{18}-3109428z^{17}+368233z^{16}+2734116z^{15}+1243978z^{14}-741000z^{13}-805907z^{12}+75696z^{11}+164040z^{10}+82560z^9-172194z^8+53440z^7-10290z^6+32440z^5-7460z^4-4400z^3+100z^2+300z+25 = 0.$$

The coefficients are particially calculated using the Mathcad package, and $\mathcal{polyroots}()$ function is also used, which calculates all the roots of the polynomial by the "accompanying matrix" method.

Calculating values $y$ with $(2)$ and $x,f$ by the formula $$x = 1-y^2-z^3,\quad f=xy^2+yz^2+zx^2$$ and checking them by substituting in the original system, we obtain the following stationary points with the positive coordinates: $$ (x,y,z,f)\in\left[\genfrac{}{}{0pt}{0} {(0.5581125,\ 0.4984120,\ 0.5783713,\ 0.4856221)} {(0.1636702,\ 0.7619816,\ 0.6347238,\ 0.4548812)} \right. $$

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  • $\begingroup$ Less time consuming way to set out as a separate answer. $\endgroup$ May 27, 2016 at 12:43
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    $\begingroup$ The problem with this is that the equations were solved using some solver (approximations) and so cant represent exact solution $\endgroup$
    – ibnAbu
    May 27, 2016 at 18:49
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    $\begingroup$ Worst there is no proof that the approximate solutions give global maximum. $\endgroup$
    – ibnAbu
    May 27, 2016 at 18:50
  • $\begingroup$ @stalker2133 Approximation seems a single way to get exact maximum, because we've got a hard task. "Classic" (clear) proof is shown there in another my answer, it was in my link. But never note it.))) $\endgroup$ May 27, 2016 at 19:44
  • $\begingroup$ I checked the link but I wonder why we are using determinant to tell about solutions of non- linear equations. $\endgroup$
    – ibnAbu
    May 27, 2016 at 21:16
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Update: A simpler prof (2023/03/24)

Since $y^2 + \frac{1}{4} \ge y$, it suffices to prove that $$x^2(y^2 + \tfrac{1}{4}) + y^2 z + z^2 x < \frac{1}{2}.$$ Since $y^2 = 1-x-z^3$, it suffices to prove that $$4x^3+(4z^3-5)x^2+(-4z^2+4z)x+4z^4-4z+2 > 0. \tag{1}$$

We have \begin{align*} &4x^3+(4z^3-5)x^2+(-4z^2+4z)x+4z^4-4z+2\\ =\ & 4x\left(x + \frac{1}{2}z^3 - \frac{5}{8}\right)^2 + \left( 4\,z-4\,{z}^{2}-{z}^{6}+\frac52\,{z}^{3}-{\frac {25}{16}} \right) x+4\,{z}^{4}-4\,z+2. \tag{2} \end{align*}

It suffices to prove that $$\left( 4\,z-4\,{z}^{2}-{z}^{6}+\frac52\,{z}^{3}-{\frac {25}{16}} \right) x+4\,{z}^{4}-4\,z+2 > 0.$$

Since $4z^4 - 4z + 2 = (2z^2 - 1)^2 + (2z - 1)^2 > 0$, we only need to prove the case that $4\,z-4\,{z}^{2}-{z}^{6}+\frac52\,{z}^{3}-{\frac {25}{16}} < 0$. Since $x + z^3 \le 1$, it suffices to prove that $$\left( 4\,z-4\,{z}^{2}-{z}^{6}+\frac52\,{z}^{3}-{\frac {25}{16}} \right) (1 - z^3) + 4\,{z}^{4}-4\,z+2 > 0$$ or $$16\,{z}^{9}-56\,{z}^{6}+64\,{z}^{5}+65\,{z}^{3}-64\,{z}^{2}+7 > 0.$$

We have \begin{align*} &16\,{z}^{9}-56\,{z}^{6}+64\,{z}^{5}+65\,{z}^{3}-64\,{z}^{2}+7\\ ={}& \frac{1}{16} ( 16{z}^{4}-1 ) ^{2}z+ \frac{7}{2}( 4{z}^{2}-1 ) ^{2} ( 1-z ) z+ \frac58( 4{z}^{2}-1 ) ^ {2}z + 7 ( 2z-1 ) ^{2} ( 1-z ) z\\[6pt] &\qquad + 42(z - 9/16)^2 z + \frac{87}{4} z^2 - \frac{3133}{128}z + 7\\ >{}& 0 \end{align*} where we use $\frac{87}{4} z^2 - \frac{3133}{128}z + 7 > 0$ for all $z \in [0, 1]$.

We are done.

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    $\begingroup$ $4z^4-4z+2=(2z^2-1)^2+(2z-1)^2$ $\endgroup$ Dec 6, 2020 at 0:09
  • $\begingroup$ @YuriNegometyanov Very nice! Thanks. $\endgroup$
    – River Li
    Dec 6, 2020 at 0:21
  • $\begingroup$ And try $18\times(1) - (6x+2z^3-5/2)\times(1)',$ this is correct in the stationary point. $\endgroup$ Dec 6, 2020 at 0:47
  • $\begingroup$ @YuriNegometyanov Thanks. Which one is $(1)'$? $\endgroup$
    – River Li
    Dec 6, 2020 at 1:17
  • $\begingroup$ Partial derivative of LHS(1) by $x.$ $\endgroup$ Dec 6, 2020 at 1:23
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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 $\dfrac{A+B+C}{4}$.

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    $\begingroup$ Isn't a+b+c=2, and that would change the answer $\endgroup$
    – avz2611
    May 24, 2016 at 15:18
  • $\begingroup$ Thanks, I'll modify the proof. $\endgroup$
    – Aravind
    May 24, 2016 at 18:00
  • 1
    $\begingroup$ so assuming those three additional inequalities, can you at least prove the requested inequality? If not what progress does this represent? $\endgroup$
    – hkBst
    May 27, 2016 at 10:28
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enter image description here

The method to be employed here is exactly the same as in a previous answer:

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).

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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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  • $\begingroup$ It looks like all we have to do is search for the farthest point away from the local minimum that also satisfies the constraint. $\endgroup$
    – ibnAbu
    May 24, 2016 at 7:58
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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 $p$.

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    $\begingroup$ You find a maximum at $p = \frac{3}{4}\left(\frac{1}{4}\right)^{\frac{1}{3}} \approx 0.47247$. However, in the comments to the question, Ewan Delanoy found a point which has value $0.484818$. So I don't think your last paragraph will work? $\endgroup$
    – sTertooy
    May 24, 2016 at 12:07
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    $\begingroup$ I was hoping someone else would address that... $\endgroup$
    – tomi
    May 24, 2016 at 12:12
  • $\begingroup$ It also seems odd that the maximum I found is at $\frac {\sqrt 3} 2$. Is there some trig transformation that would make this easier? $\endgroup$
    – tomi
    May 24, 2016 at 12:14
  • $\begingroup$ @stalker2133 look there: math.stackexchange.com/questions/1775498/… $\endgroup$ May 27, 2016 at 18:26
  • $\begingroup$ @stalker2133 There are two my answers. Follow link, please. $\endgroup$ May 27, 2016 at 19:37
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$\color{green}{\textbf{Computer free version of 17.12.20.}}$

$\color{brown}{\textbf{Preliminary notes.}}$

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!

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    $\begingroup$ 1) Should it be $f(x,0,z) = z^2x = (1-z^3)z^2 \le (1-z^3)z \le H, f(0,y,z) = y^2z = (1-z^3)z \le H$? $\endgroup$
    – River Li
    Dec 18, 2020 at 3:32
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    $\begingroup$ 2) How do you get the 3rd equation in (5)? How do you get the equation between (5) and (6)? More details should be given. $\endgroup$
    – River Li
    Dec 18, 2020 at 3:32
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    $\begingroup$ 3) Also, the 1st equation in (4) is different from RHS of (2)? $\endgroup$
    – River Li
    Dec 18, 2020 at 3:41
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    $\begingroup$ 4) Check (6) $x = \cdots$ $\endgroup$
    – River Li
    Dec 18, 2020 at 12:31
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    $\begingroup$ The sentence around "has a vertical asymptote at", maybe the details should be given how to get $h(z) > 1$? Also, the next paragraph... $\endgroup$
    – River Li
    Dec 18, 2020 at 15:35
0
$\begingroup$

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.)

$\endgroup$

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