<s>[INST] 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 . 
 [/INST] As pointed out by River Li, first show that for every $x>0$ holds the inequality
$$
\log x\leq\frac{x}{8\sqrt{3}}+\log\left(\frac{8\sqrt{3}}{e}\right).\tag 1
$$
The equality in $(1)$ for holds only for $x=8\sqrt{3}$.
From $(1)$ we have
$$
\sum_{cyc}\frac{\log(7a+b)}{7a+b}\leq \frac{\sqrt{3}}{8}+\sum_{cyc}\frac{1}{7a+b}\log\left(\frac{8\sqrt{3}}{e}\right).
$$
Set 
$$
\Pi(x,y,z):=\sum_{cyc}\frac{1}{7x+y},
$$
Consider the inequality
$$
\frac{1}{kx+ly}\leq \frac{A}{kx}+\frac{B}{ly},
$$
with $k,l,x,y>0$, $k=7$, $l=1$ and assume that
$$
\frac{A}{k}+\frac{B}{l}=\frac{1}{k+l}.
$$
Hence $B=\frac{1}{56}(7-8A)$. But for this value of $B$ it is
$$
\frac{1}{kx+ly}-\frac{A}{7x}-\frac{B}{y}=-\frac{(x-y)(49x-56Ax-8Ay)}{56xy(7x+y)}.
$$
Using $49-56A=8A\Leftrightarrow A=\frac{49}{24}$, we get
$$
\frac{1}{7x+y}-\frac{7}{64x}-\frac{1}{64y}=-\frac{7(x-y)^2}{64xy(7x+y)}\tag 2
$$
Summing all three equalities $(2)$ (for the pairs $(x,y),(y,z),(z,x)$) we get
$$
\sum_{cyc}\frac{1}{7x+y}\leq\frac{7}{64}\sum_{cyc}\frac{1}{x}+\frac{1}{64}\sum_{cyc}\frac{1}{x}
$$
and consequently
$$
\Pi(x,y,z)\leq\frac{1}{8}\sum_{cyc}\frac{1}{x}.
$$
This is a ''bad'' estimate, since 
$$
\sum_{cyc}\frac{1}{x}\geq\sqrt{3}.
$$
However, I think, someone can work better with $(2)$ and obtain a ''good'' estimate.
</s>