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.    
 A: 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!
A: A bit algebra shows that the inequality is equivalent to
$$
5 \left(5 a^5 \left(13 b^2+5 c^2\right)+13 a^4 \left(5 b^3-13 b^2 c-5 b c^2-5 c^3\right)+a^3 \left(-65 b^4+144 b^2 c^2+65 c^4\right)+a^2 \left(25 b^5-65 b^4 c+144 b^3 c^2+144 b^2 c^3-169 b c^4+65 c^5\right)-13 a \left(13 b^4 c^2+5 b^2 c^4\right)+5 b^2 c^2 \left(13 b^3+13 b^2 c-13 b c^2+5 c^3\right)\right) \ge0
$$
The left hand side is a polynomial, so this can be solved by Cylindrical Algebra Decomposition -- https://en.wikipedia.org/wiki/Cylindrical_algebraic_decomposition
The following code in Mathematica does the job.
ex1 = a^3/(13 a^2 + 5 b^2) + b^3/(13 b^2 + 5 c^2) + c^3/(5 a^2 + 13 c^2) >= 1/18 (a + b + c);
ex2 = ex1[[1]] - ex1[[2]] // Together // Numerator // Simplify;
ex3 = ForAll[{a, b, c}, And @@ {a >= 0, b >= 0, c >= 0}, ex2 >= 0];
CylindricalDecomposition[ex3, {}]

Note this may take a few minutes to run.
A: 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.
A: 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) :
$$p(0.785,1.25+x,1.861+x^n)$$
A: 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 !
