Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ 
$x,y,z >0$, prove
  $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$

Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame. 
Update 1: In this forum, somebody said  that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant. 
Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.  
Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.
Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".
 A: I write a start for a full answer (this is an idea that @Starfall first proposed in comment). If someone wants to use it to end the proof, she/he is welcome!
Let
$$f(x,y,z):=\frac{x^4}{ax^3+by^3}+\frac{y^4}{ay^3+bz^3}+\frac{z^4}{az^3+bx^3}.$$
Since $f$ is homogeneous of degree 1, it is sufficient to consider $x,y,z$ on the plane $P:=\{x+y+z=1\}$. Let
$$g(x,y,z):=x+y+z-1$$
be the constraint function. We compute :
$$\mathrm{d}f(x,y,z)=\left(\frac{ax^6+4bx^3y^3}{(ax^3+by^3)^2}-\frac{3bx^2z^4}{(az^3+bx^3)^2}\right)\mathrm{d}x+\left(\frac{ay^6+4by^3z^3}{(ay^3+bz^3)^2}-\frac{3bx^4y^2}{(ax^3+by^3)^2}\right)\mathrm{d}y$$
$$+\left(\frac{az^6+4bx^3z^3}{(az^3+bx^3)^2}-\frac{3by^4z^2}{(ay^3+bz^3)^2}\right)\mathrm{d}z,$$
$$\mathrm{d}g(x,y,z)=\mathrm{d}x+\mathrm{d}y+\mathrm{d}z.$$
Define the $2\times 3$ matrix
$$M:=\begin{pmatrix}
\frac{\partial f}{\partial x}(x,y,z) & \frac{\partial f}{\partial y}(x,y,z) & \frac{\partial f}{\partial z}(x,y,z)\\
\frac{\partial g}{\partial x}(x,y,z) & \frac{\partial g}{\partial y}(x,y,z) & \frac{\partial g}{\partial z}(x,y,z)
\end{pmatrix}.$$
By Lagrange multipliers theorem, all the 3 sub-determinants of $M$ must vanish at a local minimum $(x,y,z)$ of $f$ on $P$.
Setting
$$A:=ax^3+by^3,\quad B:=az^3+bx^3,\quad ay^3+bz^3,$$
cancelling the 3 sub-determinants of $M$ yields :
\begin{align}
\begin{cases}
B^2C^2(ax^6+4bx^3y^3+3bx^4y^2)-3A^2C^2bx^2z^4-A^2B^2(ay^6+4by^3z^3)&=0\\
B^2C^2(ax^6+4bx^3y^3)-A^2C^2(3bx^2z^4+az^6+4bx^3z^3)+3A^2B^2by^4z^2&=0\\
A^2B^2(ay^6+4by^3z^3+3by^4z^2)-3B^2C^2bx^4y^2-A^2C^2(az^6+4bx^3z^3)&=0\\
x+y+z=1,\ x,y,z>0
\end{cases}.
\end{align}
Labelling the lines $(1)$, $(2)$, $(3)$ and $(4)$, we can see that $(1)-(2)=-(3)$, so that we can forget one of the three first lines.
Here we need to do some (boring) algebra, using the constraints of the fourth line above and maybe some tricks like writing $ax^3=A-by^3$ and $bx^4=(1-y-z)(B-az^3)$. But I am too busy now to try this, and I don't know if I would try later...
A: For checking purposes.
Making $y = \lambda, \ z = \mu x$ and substituting into
$$
f(x,y,z) = \frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} - \frac{x+y+z}{13}
$$ 
giving
$$
g(x,\lambda,\mu) =x\left( \frac{1}{5 \lambda ^3+8}+\frac{\lambda ^4}{8 \lambda ^3+5 \mu ^3}+\frac{\mu ^4}{8 \mu ^3+5}-\frac{1}{13} (\lambda +\mu +1)\right)
$$
and discarding $x > 0$ we get
$$
\mathcal{G}(\lambda,\mu) = \frac{1}{5 \lambda ^3+8}+\frac{\lambda ^4}{8 \lambda ^3+5 \mu ^3}+\frac{\mu ^4}{8 \mu ^3+5}-\frac{1}{13} (\lambda +\mu +1)
$$
Now solving the stationary conditions
$$
\nabla\mathcal{G}(\lambda,\mu) = 0
$$
we have the feasible stationary points with qualification.
$$
\left[
\begin{array}{cccl}
 \lambda & \mu & \mathcal{G}(\lambda,\mu) & \mbox{kind} \\
 1. & 1. & 0. & \mbox{min} \\
 0.485435 & 0.715221 & 0.000622453 & \mbox{min}\\
 0.646265 & 0.811309 & 0.000758688 & \mbox{saddle} \\
 1.37554 & 0.688678 & 0.000863479 & \mbox{min} \\
 1.25 & 0.77611 & 0.000941355 & \mbox{saddle} \\
 1.38778 & 1.85522 & 0.00123052 & \mbox{min} \\
 1.34211 & 1.74761 & 0.00123288 & \mbox{saddle} \\
\end{array}
\right]
$$
so the best solution is at $x = y = z = 1$
Attached the level contours for $\mathcal{G}(\lambda,\mu)$ with the stationary points in red.

A: Not sure, if I missed out anything here. Take a look.
For non negative, $X,Y,Z$, 
We  can perhaps use Titu's inequality (a mix of Holder and CS), sometimes called Titu's screw lemma (https://en.wikipedia.org/wiki/Nesbitt%27s_inequality).
\begin{equation}
\sum_{k=1}^{n}{\frac{x_{k}^{2}}{a_{k}}} \ge \frac{\left(\sum_{k=1}^{n}{x_{k}}\right)^{2}}{\sum_{k=1}^{n}{a_{k}}}
\end{equation}
With $n\to3$ terms, $x_{1}\to X^{2},x_{2} \to Y^{2}, x_{3} \to Z^{2}$ and $a_{1} \to A, a_{2}\to B, a_{3} \to C$, we will have 
\begin{eqnarray*}
\frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C} &\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\
\end{eqnarray*}
With 
\begin{eqnarray*}
A &=& \alpha X^{3} +\beta Y^{3} \\
B &=& \alpha Y^{3} +\beta Z^{3} \\
C &=& \alpha Z^{3} +\beta X^{3}
\end{eqnarray*}
where,
\begin{eqnarray*}
A+B+C &=& (\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)
\end{eqnarray*}
\begin{eqnarray}
\frac{X^4}{A}+\frac{Y^4}{B}+\frac{Z^4}{C} &=&\frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C}\\
 &\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\
 &=& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)}  \\
 &\overset{(p)}{\ge}& \frac{\left(X^{3}+Y^{3}+Z^{3}\right)\left(X+Y+Z\right)}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)}  \\
  &=& \frac{\left(X+Y+Z\right)}{(\alpha+\beta)}
\end{eqnarray}
QED.
Here $(p)$ is from the fact that,
\begin{eqnarray*}
(X^2+Y^2+Z^2)^{2} -\left(X^{3}+Y^{3}+Z^{3} \right) (X+Y+Z) &=& XY(X-Y)^{2}+YZ(Y-Z)^{2}+ZX(Z-X)^{2} \\
&\ge& 0
\end{eqnarray*}
Here $\alpha=8$ and $\beta=5$.
A: Let's reform this inequality in a way such that we can comprehend it better. Define $a=\dfrac{y}{x}$ and $b=\dfrac{z}{y}$, therefore $\dfrac{x}{z}={1\over ab}$. We can suume without lose of generality that $a,b\le1$ We need to prove that $$\dfrac{x}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{y}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{z}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{x+y+z}{13}$$by dividing the two sides of the inequality by $x$ and substituting $a,b,c$ we have that$$\dfrac{1}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{\dfrac{y}{x}}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{\dfrac{z}{y}}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{1+\dfrac{y}{x}+\dfrac{z}{x}}{13}$$and $$\dfrac{1}{8+5a^3}+\dfrac{a}{8+5b^3}+\dfrac{a^4b^4}{5+8a^3b^3}\ge \dfrac{1}{13}+\dfrac{a}{13}+\dfrac{ab}{13}$$which is equivalent to $$\left(\dfrac{1}{8+5a^3}-\dfrac{1}{13}\right)+\left(\dfrac{a}{8+5b^3}-\dfrac{a}{13}\right)+\left(\dfrac{a^4b^4}{5+8a^3b^3}-\dfrac{ab}{13}\right)\ge 0$$by simplifying each of the components and multiplying both sides in $\dfrac{13}{5}$ we obtain$$\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}\ge0$$below is a depiction of $f(a,b)=\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}$ for $0\le a,b\le 1$

which proves the inequality graphically (I believe that Lagrange multipliers or any other method based on 1st order derivations may help but i hadn't much time to think on it hope you find an analytic way) but neither such a time i spent on the problem nor a computer is given us in the exam :) also i appreciate if any one updates his/her comment with such an analytical method. I'm really curious about that.....
A: 
This is a question of the symmetric type, such as listed in:

Why does Group Theory not come in here?

With a constraint $\;x+y+z=1\;$ and $\;x,y,z > 0$ . Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in:

How prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$

Our function $f$ in this case is:
$$
f(x,y,z) = \frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} - \frac{1}{13}
$$
And the minimum of that function inside the abovementioned triangle must shown to be greater or equal to zero. Due to symmetry - why oh why can it not be proved with Group Theory - an absolute minimum of the function is expected
at $(x,y,z) = (1/3,1/3,1/3)$. 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 + sqr(g/grens)*(max-min); { sqr = square ; grens = 20 ; g = 0..grens }

The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values.
Maximum and minimum values of the function are observed to be:


 0.00000000000000E+0000 < f < 4.80709198767699E-0002

The little $\color{blue}{\mbox{blue}}$ spot in the middle is where $\,0 \le f(x,y,z) < 0.00002$ .
A: This is too long to fit into a comment. I wanted to ask a question about my proof on this problem. (It might help discover another proof)

This proof has a flaw -- From $AB \ge C$ and $A \ge D$, I wrongly implied that $DB \ge C$. 
Is there a way to slightly modify it such that it can prove the statement or is it completely wrong?

Seeing that the inequality is homogeneous (meaning that the transformation
$(x, y, z) \mapsto (kx, ky, kz)$ does not change anything), it is natural to impose a constraint on it. So let us assume
without the loss of generality that $xyz=1$.
From Cauchy-Schwarz Inequality,
$$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3})\geqslant (x^2+y^2+z^2)^2$$
Since (By AM-GM) $$[8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3] = 13(x^3+y^3+z^3) \geqslant 13(3 \sqrt[3]{(xyz)^3}) = 13(3)$$
Therefore 
$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (13)(3)(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (x^2+y^2+z^2)^2$
Therefore
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{(x^2+y^2+z^2)^2}{(13)(3)}$$
Now it reamins to prove that $\frac{(x^2+y^2+z^2)^2}{(13)(3)} \geqslant \frac{x+y+z}{13}$, i.e.
$$(x^2+y^2+z^2)(x^2+y^2+z^2)\geqslant 3(x+y+z)$$
which is straightforward by AM-GM:
Notice that for all $xyz=1$
$$(x - 1)^2 + (y-1)^2 + (z - 1)^2 \ge 0$$
$$x^2 + y^2 + z^2 - 2a - 2b - 2c + 3 \ge 0$$
$$x^2 + y^2 + z^2 \ge -3 + (x + y + z) + (x + y + z)$$
But by AM-GM, $x + y + z \ge 3\sqrt[3]{xyz} = 3$. So,
$$x^2 + y^2 + z^2 \ge -3 + 3 + (x + y + z)$$
$$x^2 + y^2 + z^2 \ge  x + y + z \ge 3$$
A: I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :

$$\frac{a^4}{8a^3+5b^3}+\frac{b^4}{8b^3+5a^3}\geq \frac{a+b}{13}$$

Proof:
We have with $x=\frac{a}{b}$ :
$$\frac{x^4}{8x^3+5}+\frac{1}{8+5x^3}\geq \frac{1+x}{13}$$
Or 
$$\frac{5}{13}(x - 1)^2 (x + 1) (x^2 + x + 1) (5 x^2 - 8 x + 5)\geq 0$$
So we have (if we permute the variables $a,b,c$ and add the three inequalities ) :
$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}+\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{6.5}$$
If we have $\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$
We have :
$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq \frac{a+b+c}{13}$$
But also
$$\frac{(a-\epsilon)^4}{8(a-\epsilon)^3+5b^3}+\frac{(b)^4}{8(b)^3+5(c+\epsilon)^3}+\frac{(c+\epsilon)^4}{8(c+\epsilon)^3+5(a-\epsilon)^3}\geq \frac{a+b+c}{13}$$
If we put $a\geq c $ and $\epsilon=a-c$
We finally obtain :
$$\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{13}$$
If we have $\sum_{cyc}\frac{a^4}{8a^3+5b^3}\leq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$
The proof is the same as above . 
So all the cases are present so it's proved !
A: Too long for a comment.

The Engel form of Cauchy-Schwarz is not the right way:
$$\frac{(x^2)^2}{8x^3+5y^3}+\frac{(y^2)^2}{8y^3+5z^3}+\frac{(z^2)^2}{8z^3+5x^3} \geq \frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$
So we should prove that 
$$\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\geq\frac{x+y+z}{13}$$
which is equivalent to
$$\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}\geq x+y+z$$
but by Cauchy-Schwarz again we have $$x+y+z=\frac{(x^2)^2}{x^3}
+\frac{(y^2)^2}{y^3}+\frac{(z^2)^2}{z^3}
\geq \frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}$$
and the inequalities are in the wrong way.
