<s>[INST] 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.    
 [/INST] 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.
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