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.    

Your inequality is equivalent to :
$$\sum_{cyc}\frac{a}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2\geq \frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$\frac{a}{13b} \sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2+\frac{1}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{b}{c}))^2+\frac{c}{13b}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{c}{a}))^2\geq \frac{1+\frac{a}{b}+\frac{c}{b}}{18}$$
Now we put $\sqrt{\frac{13}{5}}\frac{a}{b}=x$, $\sqrt{\frac{13}{5}}\frac{b}{c}=y$, $\sqrt{\frac{13}{5}}\frac{c}{a}=z$, your inequality is equivalent to:
$$\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(x)^3}{(x^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(z)^2}{(z^2+1)}$$$$\geq \dfrac{1+\sqrt{\dfrac{5}{13}}x+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$
with the condition $xyz=(\sqrt{\frac{13}{5}})^3$.
We study the following function:
$$f(x)=\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(x)^3}{(x^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(z)^2}{(z^2+1)}-\dfrac{1+\sqrt{\dfrac{5}{13}}x+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$
This function is easily differentiable and the minimum is for $x=\sqrt{\frac{3\sqrt{77}}{17}-\frac{26}{17}}=\alpha$. So with the condition $xyz=(\sqrt{\frac{13}{5}})^3$ becomes $yz=\frac{(\sqrt{\frac{13}{5}})^3}{\alpha}=\beta$. So we have this inequality just with $y$:
$$\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(\alpha)^3}{(\alpha^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(\frac{\beta}{y})^2}{((\frac{\beta}{y})^2+1)}\geq\dfrac{1+\sqrt{\dfrac{5}{13}}\alpha+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$
which is easily analyzable. Done!
