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.    

Let $f,g,p,u,k,d,x,y,z>0$ such that $u\geq k$:
$$y=\left(\frac{x\left(g+u\right)p}{g+k}\right),x=\left(\frac{z\left(d+u\right)p}{d+k}\right),z=\left(\frac{y\left(f+u\right)p}{f+k}\right)$$
Then we have the functions :
$$h\left(x\right)=\frac{x}{13+\frac{5p^{2}\left(x+u\right)^{2}}{\left(x+k\right)^{2}}}$$
Wich is convex on $(0,\infty)$ and $u\ge k>0$ and $p> 0$  :
Then using Jensen's inequality we have :
$$\frac{z}{f}h(f)+\frac{x}{d}h(d)+\frac{y}{g}h(g)\geq \left(\frac{z}{f}+\frac{x}{d}+\frac{y}{g}\right)h\left(\frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}\right)$$
Now the problem is :
$$\left(\frac{z}{f}+\frac{x}{d}+\frac{y}{g}\right)h\left(\frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}\right)\geq^{?}\frac{\left(x+y+z\right)}{18}\tag{I}$$
After simplification we have the constraint :
$$p(\frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}+u)\leq \frac{\left(z+x+y\right)}{\frac{z}{f}+\frac{x}{d}+\frac{y}{g}}+k$$
we are done .
Last edit 23/04/2022 :
We have the inequality for $x\in[0,1]$ :
$$f(x)=\left(\frac{1}{13+5x^{-2}}\right)\geq g(x)=\left(\frac{1}{6}\left(\frac{1}{2+x^{-2}}+\frac{x^{2}\left(1-x\right)^{2}}{5}\right)\right)\left(1+\frac{\left(x\left(1-x\right)\right)^{2}}{16.75}\right)^{3}$$
Now we need to show under some assumptions given below:
$$p(a,b,c)=-\frac{\left(a+b+c\right)}{18}+ag\left(\frac{a}{b}\right)+bf\left(\frac{b}{c}\right)+\frac{c^{3}}{13c^{2}+5a^{2}}\geq 0$$
For that we use Buffalo's way and a constraint we have all the coefficient positives in ($x\ge 0$ and $n\geq 1$ a natural number) :
$$p(0.785,1.25+x,1.861+x^n)$$
