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".

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,
$$x^2 + y^2 + z^2 \ge -3 + 3 + (x + y + z)$$
$$x^2 + y^2 + z^2 \ge  x + y + z \ge 3$$
