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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

Chinnapparaj R
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max8128
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  • It is false for $(a,b,c)=(0.785, 1.25, 1.861)$, the example given in the referenced question https://math.stackexchange.com/q/1777075/42969. – Martin R Nov 23 '18 at 08:58

1 Answers1

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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 $$RS=\frac{5}{324}.$$

Michael Rozenberg
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