Inequality $\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$ I'm interested by the following problem :

Let $a,b,c>0$ such that $abc=a+b+c$ then we have :
  $$\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$$

I have tried to use convexity and Jensen but the result is weaker  . I try also Karamata's inequality but it fails totaly so I'm a bit lost .
If you have a hint it would be great .
Thank you . 

Using the fact that $\ln x \le \ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(x-8\sqrt{3})$ for $x > 0$
(the proof is simple and thus omitted),
it suffices to prove that
$$\sum_{\mathrm{cyc}}\frac{\ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(7a+b-8\sqrt{3})}{7a+b}
\le \frac{3\ln (8\sqrt{3})}{8\sqrt{3}}$$
or
$$\frac{1}{7a+b} + \frac{1}{7b+c} + \frac{1}{7c+a} \le \frac{\sqrt{3}}{8}.$$
This has been solved in the link below:
If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$
