Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$ 
Let $a$, $b$ and $c$ be positive numbers. Prove that:
  $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$

This inequality is strengthening of the following Vasile Cirtoaje's one, which he  created in 2005.

Let $a$, $b$ and $c$ be positive numbers. Prove that:
  $$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$

My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is  $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.    

I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :

$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5a^2}\geq \frac{a+b}{18}$$

Proof:
We have with $x=\frac{a}{b}$ :
$$\frac{x^3}{13x^2+5}+\frac{1}{13+5x^2}\geq \frac{1+x}{18}$$
Or 
$$5(x+1)(x-1)^2(5x^2-8x+5)\geq 0$$
So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :
$$\sum_{cyc}\frac{a^3}{13a^2+5b^2}+\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{9}$$
If we have $\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq\sum_{cyc}\frac{a^3}{13a^2+5c^2}$
We have :
$$\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq \frac{a+b+c}{18}$$
But also
$$\frac{(a-\epsilon)^3}{13(a-\epsilon)^2+5b^2}+\frac{(b)^3}{13(b)^2+5(c+\epsilon)^2}+\frac{(c+\epsilon)^3}{13(c+\epsilon)^2+5(a-\epsilon)^2}\geq \frac{a+b+c}{18}$$
If we put $a\geq c $ and $\epsilon=a-c$
We finally obtain :
$$\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{18}$$
So all the cases are here so it's proved !
