$\color{green}{\textbf{Version of 01.04.24. Common way.}}$
$\color{brown}{\;\textbf{1. Cubic in the specific form.}}$
The equation
$$\dfrac{3t}{4t^3+d}=c^2,\qquad c>0,\quad c^3d\in[0,1]\tag{1.1}$$
can be solved via the trigonometric form:
$$\dfrac{3ct}{4(ct)^3+c^3d}=1,\quad 3ct-4(ct)^3=c^3d,\quad
\sin(3\arcsin(ct))=c^3d,$$
$$\mathbf{t_k=\dfrac1c\,sin\;\dfrac{2\pi k+\arcsin(c^3d)}3},\qquad k=0,1,2.\tag{1.2}$$
Since $\;c^3d\in[0,1],\;$ then $\;\arcsin(c^3d)\in\left[0,\dfrac\pi2\right],\;$
$$t_0=\dfrac1c\,\sin\;\dfrac{\arcsin(c^3d)}3\in\left[0,\dfrac1{2c}\right];\tag{1.3}$$
$$t_1=\dfrac1c\,\sin\;\dfrac{2\pi+\arcsin(c^3d)}3
=\dfrac1c\,\sin\;\dfrac{\pi-\arcsin(c^3d)}3\in\left[\dfrac1{2c},\dfrac{\sqrt3}{2c}\right];\tag{1.4}$$
$$t_2=\dfrac1c\,\sin\;\dfrac{4\pi+\arcsin(c^3d)}3=-\dfrac1c\,\sin\;\dfrac{\pi+\arcsin(c^3d)}3\in\left[-\dfrac{\sqrt3}{2c},-\dfrac{1}{c}\right].\tag{1.5}$$
$\color{brown}{\textbf{2. The main areas.}}$
Taking in account the rotational symmetry of the given inequality
$$\dfrac{x^4}{8x^3+5y^3}+\dfrac{y^4}{8y^3+5z^3}+\dfrac{z^4}{8z^3+5x^3}\ge\dfrac{x+y+z}{13},\tag{2.1}$$
we can WLOG assume $\;x\;$ the greatest one of the unknowns $\;\{x,y,z\}.\;$
Then it suffices to consider two main cases (areas):
$A.\quad 0\le z\le y\le x.$
$B.\quad0\le y\le z\le x.$
$\color{brown}{\textbf{3. Edges.}}$
$\color{teal}{\textbf{3a. Edges, case A.}}$
Let $\;\color{teal}{\mathbf{\;z=0,\;0<y\le x.\;}}$
Then inequality ($2.1$) takes a form of
$$\dfrac{x^4}{8x^3+5y^3}+\dfrac{y^4}{8y^3}\ge\dfrac{x+y}{13}.$$
Substitution $u=\dfrac yx\in[0,1]\;$ presents it in the form of $\;L(u)\ge0,\;$
where
$$L(u)=\dfrac1{8+5u^3}+\dfrac18u-\dfrac{1+u}{13}
=\dfrac{104+13u(8+5u^3)-8(1+u)(8+5u^3)}{104(8+5u^3)}$$
$$=\dfrac{40+40u-40u^3+25u^4}{104(8+5u^3)}
=\dfrac5{104}\dfrac{8+8u(1-u^2)+5u^4}{8+5u^3}\ge0.$$
The case is proven.
$\color{teal}{\textbf{3b. Edges, case B.}}$
Let $\color{teal}{\mathbf{\;y=0,\;0<z\le x.\;}}$
Then inequality ($2.1$) takes a form of
$$\dfrac{x^4}{8x^3}+\dfrac{z^4}{8z^3+5x^3}\ge\dfrac{x+z}{13}.$$
Substitution $\;u=\dfrac zx\in[0,1]\;$ presents it in the form of $\;L(u)\ge0,\;$
where
$$L(u)=\dfrac18+\dfrac{u^4}{5+8u^3}-\dfrac{1+u}{13}
=\dfrac{13(5+8u^3)+104u^4-8(1+u)(5+8u^3)}{104(5+8u^3)}$$
$$=\dfrac{25-40u+40u^3+40u^4}{104(5+8u^3)}
=\dfrac{15u^4+17(1-u^2)^2+2(2-5u-2u^2)^2}{104(5+8u^3)}\ge0.$$
The case is proven.
Therefore,inequality ($2.1$) $\color{green}{\textbf{is proven at the bounds of the given area.}}$
$\color{brown}{\textbf{4. Substitutions.}}$
In the case "A", $\;\color{teal}{\mathbf{0<z\le y\le x.}}$
Substitutions
$$u=\dfrac yx\in[0,1],\quad v=\dfrac zy\in[0,1],\quad w=\dfrac xz=\dfrac1{uv}\in[1,\infty),\tag{4.1}$$
lead to presentation of the given inequality in the form of
$$\dfrac{1}{8+5u^3}-\dfrac1{13}+u\left(\dfrac{1}{8+5v^3}-\dfrac1{13}\right)+uv\left(\dfrac1{8+5w^3}-\dfrac1{13}\right)\ge 0.\tag{4.2}$$
In the case "B", $\;\color{teal}{\mathbf{0<y\le z\le x.}}$
Substitutions
$$u=\dfrac zy\in[1,\infty],\quad v=\dfrac xz\in[1,\infty],\quad w=\dfrac yx=\dfrac1{uv}\in[0,1],\tag{4.3}$$
lead to presentation of the given inequality in the form of
$$\dfrac{1}{8+5w^3}-\dfrac1{13}+\dfrac1{uv}\left(\dfrac{1}{8+5u^3}-\dfrac1{13}\right)+\dfrac1v\,\left(\dfrac1{8+5v^3}-\dfrac1{13}\right)\ge0.\tag{4.4}$$
Both ($\text{4.2}$) and ($\text{4.4}$) can be written in the form of
$$\dfrac1{uv}\,\dfrac{1-u^3}{8+5u^3}+\dfrac1v\,\dfrac{1-v^3}{8+5v^3}+\dfrac{1-w^3}{8+5w^3}\ge 0.\tag{4.5}$$
At the same time, the constraints ($\text{4.1}$) and ($\text{4.3}$) are specific ones.
Let
$$\begin{cases}
P(t)=\dfrac{1-t^3}{8+5t^3},\\[4pt]
Q(t)=-P'(t)=\dfrac{39t^2}{(8+5t^3)^2},\\[4pt]
R(t)=P(t)+tQ(t)=\dfrac{8+36t^3-5t^6}{(8+5t^3)^2},\tag{4.6}
\end{cases}$$
then (${4.5}$) takes the form of $\;f(u,v)\ge0,\;$ where
$$f(u,v)=P\left(\dfrac1{uv}\right)+\dfrac1v\left(P(v)+\dfrac1u\,P(u)\right).\tag{4.7}$$
$\color{brown}{\textbf{5. Inner stationary points.}}$
$\color{teal}{\textbf{5.1. Algebraic system.}}$
Stationary points can be defined from the system $\;f'_u=f'_v=0,\;$ or
$$\begin{cases}
P'\left(\dfrac1{uv}\right)\left(-\dfrac1{u^2v}\right)+\dfrac1v\left(-\dfrac1{u^2}\right)P(u)+\dfrac1{uv}\,P'(u)=0\\[4pt]
P'\left(\dfrac1{uv}\right)\left(-\dfrac1{uv^2}\right)-\dfrac1{v^2}\left(P(v)+\dfrac1uP(u)\right)+\dfrac1vP'(v)=0,
\end{cases}$$
$$\begin{cases}
Q(w)-P(u)-uQ(u)=0\\[4pt]
Q(w)-P(u)-uP(v)-uvQ(v)=0\\[4pt]
uvw=1
\end{cases}\Rightarrow
\begin{cases}
R(u)=Q(w)\\[4pt]
R(v)=Q(u)\\[4pt]
uvw=1
\end{cases}\tag{5.1}$$
$\color{teal}{\textbf{5.2. The inequality transformation.}}$
Taking in account the equalities $(4.6),(4.7),(5.1)$ in the form of
$$P(t)=R(t)-tQ(t),\quad R(u)=Q(w),\quad R(v)=Q(u),\quad uvw=1,$$
the inequality
$$wf(u,v,w)=P(w)+w(P(u)+uP(v))\ge0$$
can be presented in a form of
$$R(w)-wQ(w)+w\big(R(u)-uQ(u)+uR(v)-uvQ(v)\big)\ge0,$$
$$R(w)+w\big(R(u)-Q(w)\big)+uw\big(R(v)-Q(u)\big)-uvwQ(v)\ge0,$$
or
$$R(w)-Q(v)\ge0.\tag{5.2}$$
$\color{brown}{\textbf{6. Explicit form of the inverse functions.}}$
$\color{teal}{\textbf{6.1. Inversion of $Q(t).$}}$
Function $Q(t)$ can be presented in the form of
$$Q(t)= \dfrac{39t^2}{(8+5t^3)^2}=\dfrac{208}{75}\left(\dfrac{3t}{\dfrac{32}5+4t^3}\right)^2.\tag{6.1}$$
![Q(t)](https://cdn.statically.io/img/i.sstatic.net/W8OmR.png)
Then
- $Q(0)=0;$
- $\;\lim\limits_{t\to\infty}Q(t)=0;$
- $\;\dfrac{\text d}{\text dt}\dfrac{t}{8+5t^3}=\dfrac{2(4-5t^3)}{(8+5t^3)^2};\;$
- $\max Q(t)=Q_m=\dfrac{13\sqrt[\large3]{10}}{120}\approx0.233397091\;$at$\;t_m=\sqrt[\large3]{\dfrac45}\approx0.928317767.$
- $Q(t)\;$ increases at $\;t\in[0,t_m];$
- $Q(t)\;$ decreases at $\;t\in[t_m,\infty].$
Taking in account $(1.1)-(1.2),$ inverse function $\;Q^{-1}(q)\;$ finally is
$$\theta(k,q)=2\sqrt[\large4]{\dfrac{13}{75q}} \sin\left(\dfrac13\left(2\pi k+\arcsin\left(\dfrac{4}5\left(\dfrac{75q}{13}\right)^{\large\frac34}\right)\right)\right),\quad k=0,1.\tag{6.2}$$
Note that, in accordance with $(1.5),\quad\theta(2,q)<0.$
Therefore, there are two branches of the inverse function $Q^{-1}(q):$
- "zero" branch $\;\theta(0,q),\quad q\in\left(0,Q_m\right],\quad\theta(0,q)\in[0,t_m],\;$ and
- "first" branch $\;\theta(1,q),\quad q\in\left(0,Q_m\right],\quad \theta(1,q)\in[t_m,\infty].$
Easily to check,that
- $Q(\theta(k,q))=q,\quad q\in(0,\infty);$
- $\theta(0,Q(t))=t,\quad t\in[0,t_m];$
- $\theta(1,Q(t))=t,\quad t\in[t_m,\infty].$
$\color{teal}{\textbf{6.2. Inversion of $R(t).$}}$
Function $R(t)$ can be presented in the form of
$$R(t)=\dfrac{8+36t^3-5t^6}{(8+5t^3)^2}=\dfrac7{30}-\dfrac{13}{30}\left(\dfrac{12}{8+5t^3}-1\right)^2.\tag{6.3}$$
![r(t)](https://cdn.statically.io/img/i.sstatic.net/vqm2C.png)
Easily to see, that
- $R(0)=\dfrac18=0.125;$
- the positive root of $R(t)$ is
$\;t_h=\sqrt[\large3]{\dfrac25\,\left(9+\sqrt{91}\right)} \approx1.950077297;$
- $\max R(t)=R_m=\dfrac7{30}\approx0.233333333\;$ at $\;t_m=\sqrt[\large3]{\dfrac45}\approx0.928317767;$
- $R(t)\;$ increases at $\;t\in[0,t_m];$
- $R(t)\;$ decreases at $\;t\in[t_m,t_h];$
- $R(t)\;$ is negative at $\;t\in(t_h,\infty).$
Finally, the inverse function is
$$\rho(s,r)=2\left(\dfrac{0.3}{1+s\sqrt{\dfrac{7-30r}{13}}}-0.2\right)^{\large\frac13},\tag{6.4}$$
where $\;s=\pm1.$
Accordingly, there are two branches of the inverse function $R^{-1}(r):$
- "plus"-branch $\;\rho(1,r),\quad r\in\left[\dfrac18,\dfrac7{30}\right], \quad\rho(1,r)\in[0,t_m],\;$ and
- "minus"-branch $\;\rho(-1,r),\quad r\in\left[0,\dfrac7{30}\right], \quad\rho(-1,r)\in[t_m,t_h].$
Easily to check, that
- $R(\rho(\pm1,r))=r,\quad r\in[0,\infty);$
- $\rho(1,R(t))=t,\quad t\in[0,t_m];$
- $\rho(-1,R(t))=t,\quad t\in[t_m,t_h].$
$\color{teal}{\textbf{6.3. Explicit expressions.}}$
Since
$$Q(t)\in(0,Q_m),\quad R(t)\in(0,R_m),\quad R_m=\dfrac7{30}<Q_m,$$
then from $(5.1)$ should
$$Q(t)\in\left(0,\dfrac7{30}\right],\quad t\in\left((0,t_0]\cup[t_1,t_h]\right),\tag{6.5}$$
where
$$\begin{cases}
t_0=\theta\left(0,\dfrac7{30}\right]\approx0.917509714,\\[4pt]
t_m=\sqrt[\large3]{\dfrac45}\approx0.928317767,\\[4pt]
t_1=\theta\left(1,\dfrac7{30}\right]\approx0.939210366,\\[4pt] t_h=\rho(-1,0)\approx1.950077297.\end{cases}
\tag{6.6}$$
From $\;(5.1)\;R(u)=Q(w)\;$ should $\;u=\rho(s,Q(w))=g(s,w),\;$ where
$$g(s,t)=2\left(\dfrac{0.3}{1+s\sqrt{\dfrac7{13}-\dfrac{90t^2}{(8+5t^3)^2}}}-0.2\right)^{\large\frac13},\quad s=-1,1.\tag{6.7}$$
Also, from $\;(5.1)\;R(v)=Q(u)\;$ should $v=g(s,u).$
On the other hand, from $\;(5.1)\;R(u)=Q(w)\;$ should $w=\theta(k,R(u))=h(k,u),\;$ where
$$h(k,t)=2\sqrt[\large4]{\dfrac{13}{75R(t)}} \sin\left(\dfrac13\left(2\pi k+\arcsin\left(\dfrac{4}5\left(\dfrac{75R(t)}{13}\right) ^{\large\frac34}\right)\right)\right),\quad k=0,1.\tag{6.8}$$
Also,from $\;(5.1)\;R(v)=Q(u)\;$ should $u=h(k,v).$
At the same time,in accordance with $(5.1)$ should
$$ug(s,u)h(k,u)=1.\tag{6.9}$$
$\color{brown}{\textbf{7. Proof, case A.}}$
In accordance with $(4.1),\quad u,v\in(0,1],\quad w\in[1,t_h],\quad $ and then should
$$\;R(u)\in\left(\dfrac18,\dfrac7{30}\right],\tag{7.1}$$
$$u w\ge1.\tag{7.2}$$
$\color{teal}{\textbf{7.1. Case A,$\;u\in[0,t_0],\;k=0.$}}$
Taking in account $(1.3),(7.1),(7.2),$ easily to get
$$w=h(0,u)=\dfrac1{c(u)}\,\sin\left(\dfrac13\,\arcsin\left(\dfrac{32}5\,c^3(u)\right)\right),\tag{7.3}$$
where
$$с(u)=\sqrt[\large4]{\dfrac{75R(u)}{208}}\in\left[\sqrt[\large4]{\dfrac{75}{208}\,\dfrac18},\sqrt[\large4]{\dfrac{75}{208}\dfrac7{30}}\right]\subset[0.460762350,0.538571886],\tag{7.4}$$
$$\dfrac1{c(u)}\le\sqrt[\large4]{\dfrac{208\cdot8}{75}}<2.170316217,$$
$$\dfrac{32c^3(u)}5\le\dfrac{32}5\left(\dfrac{75}{208}\cdot\dfrac7{30}\right)^{\large\frac34}<0.999795113,$$
$$uh(0,u)<t_0\cdot\dfrac1{c(u)}\cdot\dfrac12<1.$$
and this contradicts with $(7.2).$
The plot of $\;uh(0,u)\;$ is shown below.
![Plot of uh(0,u)](https://cdn.statically.io/img/i.sstatic.net/s7Qnv.png)
$\color{teal}{\textbf{7.2. Case A,$\;u\in[0,t_0],\;k=1.$}}$
Taking in account $(1.4),$ one can get
$$\begin{align}
&w=h(1,u)=\sqrt[\large4]{\dfrac{208}{75}\dfrac{(8+5u^3)^2}{8+36u^3-5u^6}}\\[4pt]&\times\sin\left(\dfrac13\left(\pi-\arcsin\left(\dfrac45\left(\dfrac{75}{13}\dfrac{8+36u^3-5u^6}{(8+5u^3)^2}\right)^{\large\frac34} \right)\right)\right),
\end{align}\tag{7.5}$$
wherein both of multipliers are decreasing functions of $u.$
Let $\;H(t)=\dfrac1{h(1,t)}.$
Taking in account $(7.2)$ in the form of $\;u\ge\dfrac1w,\;$ one can get:
- for $\;u\in[0, H(0)]\qquad\quad\;\;\, u h(1,u) \le 1,\qquad H(0)\approx 0.629192687,$
- for $\;u\in[H(0), H(H(0))]\quad u h(1,u) \le 1,\qquad H(H(0))\approx 0.805791121,$
- for $\;u\in[H(H(0)),t_0]\qquad\; u h(1,u) \le 1,\qquad t_0\approx 0.917509714.$
Therefore,
- for $\;u\in[0,t_0]\quad u h(1,u) \le 1.$
The plot of $\;uh(1,u)\;$ is shown below.
![Plot of uW(1,u)](https://cdn.statically.io/img/i.sstatic.net/1gU3Z.png)
$\color{teal}{\textbf{7.3. Case A,$\;u\in(t_0,t_1).$}}$
If $\;u\in(t_0,t_1),\;$ then $\;Q(u)\in\left(\dfrac7{30},\dfrac{13\sqrt[\large3]{10}}{120}\right).$
At the same time, from $(5.1)$ should the equation
$$Q(u)=R(v),\quad R(v)\in\left[0,\dfrac7{30}\right],$$
which has not real solutions.
$\color{teal}{\textbf{7.4. Case A,$\;u\in[t_1,1].$}}$
Since $\;R(u)\in\left(\dfrac3{13},\dfrac7{30}\right],\;$ then
$\;w=h(0,u)\in[0.860147050,0.917509715],\;$
in contradiction with the conditions $(4.1).$
$\;(w=h(1,u)\in[0.943496443,1])\wedge(w\in[1,\infty)),$
with the solution $\;\color{green}{\mathbf{\{u,v,w\}=\{1,1,1\}}},\;$
which corresponds with the inequality $(4.5).$
Therefore, the inequality $(4.5)\;\color{green}{\textbf{is proven in the case A}}.$
$\color{brown}{\textbf{8. Proof, case B,$\;u\in[1,t_h]$.}}$
$\color{teal}{\textbf{8.1. Preliminary notes.}}$
Taking in account $(4.3),$ should $\;u\in(1,t_h),\quad v\in(1,\infty),\quad w\in(0,1),\quad t_h=\rho(-1,0).$
Then
$$\color{teal}{\mathbf{v=g(-1,u),\quad w=h(0,u)}}.\tag{8.1}$$
Graphic analysis of the functions $\;h(0,u)\;$ (orange line) and $\;u g(-1,u)\;$ (green line)
![ug(-1,u), h(0,u)](https://cdn.statically.io/img/i.sstatic.net/ew0Gr.png)
show, that both of them allow accurate linear approximations at the interval $\;u\in(1, t_h):$
$$h(0,u)\approx 0.860147-0.796(u-1)\pm 0.0133,\quad
ug(-1,u)\approx 1+2.3(u-1)\pm 0.04.$$
So their production can be approximated via quadratic function.
Therefore, the function
$B(u)= u g(-1,u) h(0,u)$ is unimodal one in the interval $u\in(1,t_h).$
$\color{teal}{\textbf{8.2. Proof.}}$
From $(8.1),(6.7),(6.8)$ should that
$$B(u)=ug(-1,u)h(0,u)=1\tag{8.2}$$
in the stationary points of the $f(u,v).$
Since $B(u)$ has the single maximum at $u\in(1,t_h),$ then equation $(8.2)$ can not have more than two roots in this interval.
Let us calculate the table
$$\hspace{-32mu}\begin{vmatrix}
u & v=g(-1,u) & w=\theta(0,R(u)) & uvw & R(w) & Q(v)\\
\!1.25&\!1.228103035&\!0.650433454&\!0.998499124&\!0.199388943&\!0.197417010\\
\!1.26&\!1.236402046&\!0.642696247&\!1.001235004&\!0.197754855&\!0.195782736\\
\!1.47&\!1.395850784&\!0.487833698&\!1.000986283&\!0.164510886&\!0.162892269\\
\!1.48&\!1.402726182&\!0.480700601&\!0.997951152&\!0.163086876&\!0.161467543\tag{8.3}
\end{vmatrix}$$
Easily to see that there are two solutions of $(7.4)$,which are situated at the intervals $(1.25,1.26)$ and $(1.47,1.48)$ accordingly.
At the same time, the functions $R(w)$ and $Q(v)$ are monotonic in these intervals. In particular, in the interval $(1.25,1.26)\quad R(w)>0.1977\;$ and $\;Q(v)<0.1975.$ This means that the inequality $(7.2)$ is satisfied at the interval $\;u\in(1.25,1.26),$ including the stationary point.
Similarly, at the interval $(1.47,1.48)$ easily to get $\;R(w)>0.1630>0.1629>Q(w).$ This means that the inequality $(7.2)$ is satisfied at the interval $\;u\in(1.47,1.48),$ including the stationary point.
Plot of the equality $uvw=1$ see below.
![Plotuvw=1](https://cdn.statically.io/img/i.sstatic.net/1zzTZ.png)
Therefore, the inequality $(4.5)$ $\color{green}{\textbf{is satisfied in the case B}}.$
$\color{green}{\textbf{Proved!!!}}$