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It's related to this If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$:

In fact we have this refinement (wich I think much easier) :

Let $a,b,c>0$ then we have : $$\frac{2}{3\sqrt{3}}\Big(\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\Big)\Big(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\Big)\leq \frac{\sqrt{3}}{8}$$

If we use brut force we get the following polynomial (after eliminated a variable): $$-(329 b^4 c^2 + 168 b^4 c - 49 b^4 - 408 b^3 c^3 + 120 b^3 c^2 + 120 b^3 c - 408 b^3 - 49 b^2 c^4 + 120 b^2 c^3 - 840 b^2 c^2 + 120 b^2 c + 329 b^2 + 168 b c^4 + 120 b c^3 + 120 b c^2 + 168 b c + 329 c^4 - 408 c^3 - 49 c^2)$$

And maybe conclude with Muirhead inequality but I'm a bit lost...

Any helps is appreciated or hints .

Thanks in advance.

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It's wrong.

Try $a=b=1$ and $c=\frac{1}{10}$.

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