<s>[INST] 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".
 [/INST] 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 $\beta=5$.
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