<s>[INST] An inequality to finalize a proof Related to this Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$ and my second answer I have to prove this :

Let $a,b,c$ be real positive numbers then we have : $$\sum_{cyc}\left(\frac{a^3}{13a^2+5b^2}\right)\left(\frac{b^3}{13b^2+5c^2}\right)\geq \frac{ab+bc+ca}{18^2}$$

My try :
With Chebychev's inequality we get :
$$\sum_{cyc}\left(\frac{a^3}{13a^2+5b^2}\right)\left(\frac{b^3}{13b^2+5c^2}\right)\geq \left(\frac{ab+bc+ca}{3}\right)\left(\sum_{cyc}\left(\frac{a^2}{13a^2+5b^2}\right)\left(\frac{b^2}{13b^2+5c^2}\right)\right) $$
Remains to prove :
$$\sum_{cyc}\left(\frac{a^2}{13a^2+5b^2}\right)\left(\frac{b^2}{13b^2+5c^2}\right)\geq \frac{3}{18^2}$$
Or with the right substitution :
$$\sum_{cyc}\left(\frac{x}{13x+5y}\right)\left(\frac{y}{13y+5z}\right)\geq \frac{3}{18^2}$$
And after this I don't know what to do...
Can someone help me ? 
Thanks 
 [/INST] It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=\frac{1}{18\cdot33}+\frac{8}{33\cdot57}+\frac{8}{57\cdot18}=\frac{155}{11286}$$ and
$$RS=\frac{5}{324}.$$
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