Cfrgtkky

Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5c^2}+frac{c^3}{13c^2+5a^2}geqfrac{a+b+c}{18}$$

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:
$$frac{a^3}{2a^2+b^2}+frac{b^3}{2b^2+c^2}+frac{c^3}{2c^2+a^2}geqfrac{a+b+c}{3}.$$

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 :
$$sum_{cyc}frac{a}{13}sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2geq frac{a+b+c}{18}$$
Each side is divided by $b$, We get:
$$frac{a}{13b} sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2+frac{1}{13}sin(arctan(sqrt{frac{13}{5}}frac{b}{c}))^2+frac{c}{13b}sin(arctan(sqrt{frac{13}{5}}frac{c}{a}))^2geq frac{1+frac{a}{b}+frac{c}{b}}{18}$$
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 $x=sqrt{frac{3sqrt{77}}{17}-frac{26}{17}}=alpha$. 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 $y$:
$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(alpha)^3}{(alpha^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(frac{beta}{y})^2}{((frac{beta}{y})^2+1)}geqdfrac{1+sqrt{dfrac{5}{13}}alpha+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$
which is easily analyzable. Done!

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 $sum_{cyc}frac{a^3}{13a^2+5b^2}geqsum_{cyc}frac{a^3}{13a^2+5c^2}$

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 $ageq 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 !