<s>[INST] If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt3}{8}$$
I tried C-S:
$$\left(\sum_{cyc}\frac{1}{7a+b}\right)^2\leq\sum_{cyc}\frac{1}{(ka+mb+c)(7a+b)^2}\sum_{cyc}(ka+mb+c)=$$
$$=\sum_{cyc}\frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{k+m+1}{(ka+mb+c)(7a+b)^2}\leq\frac{3}{64abc},$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.
If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.  
Thank you! 
 [/INST] Not a proof just a simplification for the proof wich already exists :
We prove the following inequality :

Let $a,b,c>0$ such that $abc=a+b+c$ and $a\geq b \geq c$ then we have :
  $$\sum_{cyc}\frac{1}{7a+b}\leq \sum_{cyc}\frac{1}{7b+a}$$

Proof :
We work with the following equivalent :

Let $a,b,c>0$  and $a\geq b \geq c$ then we have :
  $$\sqrt{\frac{abc}{a+b+c}}\sum_{cyc}\frac{1}{7a+b}\leq \sqrt{\frac{abc}{a+b+c}}\sum_{cyc}\frac{1}{7b+a}$$

Remains to show :

Let $a,b,c>0$  and $a\geq b \geq c$ then we have :
  $$\sum_{cyc}\frac{1}{7a+b}\leq \sum_{cyc}\frac{1}{7b+a}$$

We make the difference we get :
$$\sum_{cyc}\frac{1}{7a+b}-\sum_{cyc}\frac{1}{7b+a}=-\frac{(42 (a - b) (a - c) (b - c) (7 a^2 + 57 a b + 57 a c + 7 b^2 + 57 b c + 7 c^2))}{((7 a + b) (a + 7 b) (7 a + c) (a + 7 c) (7 b + c) (b + 7 c))}\leq 0$$
I think it simplify a part of the proof of River Li and maybe create a new approach to solve the initial inequality. 
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