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 am not a reputable source, but I think I can prove the following theorem: $$\sum_{cyc}\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$
PROOF
We will need firstly the following
Lemma 1.
$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Proof.
Applying the Rearrangement inequality on $a_{1},a_{2},...,a_{n}$ and $a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$, we have that $\sum_{cyc}a_{1}a_{2}^{\alpha}$ is maximized when $a_{1},a_{2},...,a_{n}$ and $a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$ are similarly sorted. 
Therefore, we can affirm that
$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Other hand, we need the following
Lemma 2.
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
Proof.
We establish that
$$\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}+\frac{n}{m}=\frac{a_{1}}{k_{1}}$$
Operating, we get that
$$\frac{a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}{m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}=\frac{a_{1}}{k_{1}}$$
$$k_{1}\left(a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)\right)=a_{1}m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)$$
$$k_{1}a_{1}^{\alpha+1}m+k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{1}a_{1}^{\alpha}+a_{1}mk_{2}a_{2}^{\alpha}$$
$$k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{2}a_{2}^{\alpha}$$
$$\frac{n}{m}=\left(\frac{k_{2}}{k_{1}}\right)\frac{a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}$$
Therefore, we have that
$$\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{a_{1}}{k_{1}}$$
And subsequently, repeating the process for each variable, we get that
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
Now, we are ready to prove the theorem.
Applying Lemma 1, we derive that
$$\left(\frac{k_{2}}{k_{1}}\right)\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\left(\frac{k_{2}}{k_{1}}\right)\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Therefore, substituting in the expression of Lemma 2 and operating, we have that
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{\left(\frac{k_{1}}{k_{1}}\right)a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)k_{1}}$$
$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$
As we wanted to prove.
The particular case proposed follows from the direct application of the theorem.
