Let $\sqrt{3}<p\leq x\leq q\leq 8\sqrt{3}$ and $\sqrt{3}< a\leq 8\sqrt{3}$ and the constraint $(C)$ then we have :
$$f(x)+h(x)+g(x)\leq \left(\frac{f(p)-f(q)}{p-q}+\frac{g(p)-g(q)}{p-q}\right)(x-p)+h(x)+f(p)+g(p)\leq h(x)+f(p)+g(p)\leq\frac{\sqrt{3}}{8}$$
Where :
$f(x)=\frac{1}{7a+x}$$\quad$$g(x)=\frac{1}{7x+\frac{x+a}{xa-1}}$$\quad$$h(x)=\frac{1}{7\frac{x+a}{xa-1}+a}$ .
To prove the first inequality I use convexity remarking : the function $f(x)$ is obviously convex and using derivative it's also true for $g(x)$ on the interval $(\sqrt{3}, 8\sqrt{3}]$ . So it's a classical result using a chord of a convex function .
For the second we have :
$\left(\frac{f(p)-f(q)}{p-q}+\frac{g(p)-g(q)}{p-q}\right)(x-p)\leq 0$
And even :
$\frac{f(p)-f(q)}{p-q}\leq 0$$\quad$ and $\quad$$\frac{g(p)-g(q)}{p-q}\leq 0$
I don't prove it here because it's not hard .
The third can be seen as a linear function and solving the inequality in $x$ we got :
$$p\leq x\leq q=\frac{(2(-147\sqrt{3}a^3p^2+168a^3p-21\sqrt{3}a^3-21\sqrt{3}a^2p^3-4a^2p^2+123\sqrt{3}a^2p- 172a^2-28ap^3+18\sqrt{3}ap^2-4ap+24p^2))}{(49\sqrt{3}a^4p^2-56a^4p+7\sqrt{3}a^4+7\sqrt{3}a^3p^3-456a^3p^2 -41\sqrt{3}a^3p-8a^3-56a^2p^3+337\sqrt{3}a^2p^2-8a^2p+49\sqrt{3}a^2+49\sqrt{3}ap^3-400ap^2-287\sqrt{3}ap +336a-42\sqrt{3}p^2+392p)}\quad (C)$$
I give you the Wolfram alpha link if there is mistakes .
To finish we need to show that the interval where there is $x$ is valid and I think we can use Buffalo's way . I'm stuck here...
How to show that the constraint $C$ is valid or exists ?
Thanks in advance .
Source :
If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$
Update 06/03/2021 :
Graphicaly speaking we have :
$$q=f(a)=\frac{(2(-147\sqrt{3}a^3p^2+168a^3p-21\sqrt{3}a^3-21\sqrt{3}a^2p^3-4a^2p^2+123\sqrt{3}a^2p- 172a^2-28ap^3+18\sqrt{3}ap^2-4ap+24p^2))}{(49\sqrt{3}a^4p^2-56a^4p+7\sqrt{3}a^4+7\sqrt{3}a^3p^3-456a^3p^2 -41\sqrt{3}a^3p-8a^3-56a^2p^3+337\sqrt{3}a^2p^2-8a^2p+49\sqrt{3}a^2+49\sqrt{3}ap^3-400ap^2-287\sqrt{3}ap +336a-42\sqrt{3}p^2+392p)}$$
Is increasing (I cannot prove it) for $a\geq\sqrt{3}$ so remains to show when :
$$f(\sqrt{3})> p$$