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! 
 A: 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. 
A: This inequality is something hard
$  a\geq b\geq\sqrt{3}\geq c>0$  such that $abc=a+b+c$ and constraints $(C)$ and $(2)$ and $(3)$  then we have :
$$\frac{1}{7b+a}+\frac{1}{7a+c}+\frac{1}{7c+b}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$
And :
$$\frac{1}{7b+a}\leq \left(\frac{1}{b}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (C)$$
$$\frac{1}{7c+b}+\frac{1}{7a+c}\leq \left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (2)$$
Simplify the condition $(C)$ is easy comparing the function :
$$f(x)=\frac{1}{7x+1}$$
Or:
$$f\Big(\frac{6+\frac{a}{b}}{7}\Big)\leq f\left(\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}\right)$$
For the constraint $(2)$ we know that :
$$f(a)=\frac{1}{7c+b}+\frac{1}{7a+c}-\left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$
is an increasing function (as the difference of a positive decreasing function and a negative increasing function)  on $(\sqrt{3},\frac{\frac{3}{2}\sqrt{3}-\frac{1}{2}c+c}{c(\frac{3}{2}\sqrt{3}-\frac{1}{2}c)-1})$ with :
$$b=\frac{a+c}{ac-1}$$
So we have to find an approximation of the roots !
We can choose $b\geq\frac{3}{2}\sqrt{3}-\frac{1}{2}c\quad (3)$
A bit of algebra and we get a polynomial with a root equal to $\sqrt{3}$ and one other root .Now the magic key is,the inequality and the constraints $(C),(2)$ are homogenous so we can introduce a coefficient and play with it keeping in mind the others constraints.
I continue later...Thanks!
