For me really it's a nightmarish refinement :
Claim :
Let $a,b,c>0$ such that $abc=a+b+c$ then we have :
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{1+7\left(\frac{1}{a\sqrt{3}}+\frac{1}{b\sqrt{3}}+\frac{1}{c\sqrt{3}}\right)^{\frac{6}{5}}}$$
My attempt :
I have tried to make the problem more symmetric and I build the inequality :
$$\sum_{cyc}\left(\frac{1}{7a+b}+\frac{1}{7b+a}\right)\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+7\left(\frac{1}{a\sqrt{3}}+\frac{1}{b\sqrt{3}}+\frac{1}{c\sqrt{3}}\right)^{\frac{6}{5}}}+\frac{1}{1+7\left(\frac{1}{a\sqrt{3}}+\frac{1}{b\sqrt{3}}+\frac{1}{c\sqrt{3}}\right)^{\frac{7}{4}}}\right)$$
Using uvw's method with $u=abc=a+b+c$ and $ab+bc+ca=v$ we get :
$$\frac{8(7u^2+43v)(7uv+65u)}{1764u^4+343u^2v^2+6622u^2v+79507u^2+1764v^3}\leq \left(\frac{v}{u}\right)\left(\frac{1}{1+7\left(\frac{v}{\sqrt{3}u}\right)^{\frac{6}{5}}}+\frac{1}{1+7\left(\frac{v}{\sqrt{3}u}\right)^{\frac{7}{4}}}\right)\quad (I)$$
But I'm stuck here because it's wrong...
Other attempt :
Let $a\geq b\geq c>0$ such that $abc=a+b+c$ then we have :
$$\frac{1}{3}\sum_{cyc}\left(\frac{1}{7a+b}+\frac{1}{7b+a}-\frac{4}{13a+a+b+c}\right)\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+7\left(\frac{1}{a\sqrt{3}}+\frac{1}{b\sqrt{3}}+\frac{1}{c\sqrt{3}}\right)^{\frac{6}{5}}}\right)-\left(\frac{1}{7b+a}+\frac{1}{7a+c}+\frac{1}{7c+b}\right)$$
Where I use convexity .I cannot prove it and I think it's wrong too !
Last attempt :
$a,b,c>0$ such that $abc=a+b+c$ then we have :
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$
The problem is : it seems too good to be true even if I made some numerical check.The good news is : there is the homogeneity so maybe this inequality is true in some intervals !
How to show the claim ?
Thanks in advance !
Source : If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$