Prove that $(ab+ac+bc)\sum\limits_{cyc}\frac{1}{(a-7b)^2}\geq\frac{1}{4}$ 
Let $a$, $b$ and $c$ be non-negative numbers such that $\prod\limits_{cyc}(a-7b)\neq0$. Prove that:
  $$(ab+ac+bc)\left(\frac{1}{(a-7b)^2}+\frac{1}{(b-7c)^2}+\frac{1}{(c-7a)^2}\right)\geq\frac{1}{4}$$

I think this inequality is very interesting 
because it's similar to the known Ji Chen's inequality (Iran 1996):
$$(ab+ac+bc)\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right)\geq\frac{9}{4}$$
Example of my trying.
BW does not help:
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, we need to prove that:
$$44064(u^2-uv+v^2)a^4+864(38u^3+17u^2v+73uv^2+38v^3)a^3-$$
$$-24(217u^4-1478u^3v-2157u^2v^2-3494uv^3+217v^4)a^2+$$
$$+4(98u^5-1631u^4v+3938u^3v^2+15698u^2v^3-1463uv^4+98v^5)a+$$
$$+uv(196u^4-2793u^3v+10490u^2v^2-2457uv^3+196v^4)\geq0,$$
which is nothing. 
Thank you!

Hints:
1)Put : $A=a$,
$AB=b$,
$AC=c$
With this following substitution you can eliminate a variable
We obtain $$(B+C+BC)(\frac{1}{(1-7B)^2}+\frac{1}{(B-7C)^2}+\frac{1}{(C-7)^2})$$
2)Try to prove this :
$(x+\alpha+\frac{1}{x+\beta}+\frac{x+\alpha}{x+\beta})(\frac{1}{(1-7(x+\alpha))^2}+\frac{1}{((x+\alpha)-7\frac{1}{x+\beta})^2}+\frac{1}{(\frac{1}{x+\beta}-7)^2})$ 
$\geq  (x+\alpha+\frac{1}{x+\alpha}+1)(\frac{1}{(1-7(x+\alpha))^2}+\frac{1}{((x+\alpha)-7\frac{1}{x+\alpha})^2}+\frac{1}{(\frac{1}{x+\alpha}-7)^2}) $
With $0\leq \beta\leq \alpha \leq x$
For the last inequality it's easy to see that $$(x+\alpha+\frac{1}{x+\alpha}+1)(\frac{1}{(1-7(x+\alpha))^2}+\frac{1}{((x+\alpha)-7\frac{1}{x+\alpha})^2}+\frac{1}{(\frac{1}{x+\alpha}-7)^2}) $$
Is a translation of the function :
$$(x+\frac{1}{x}+1)(\frac{1}{(1-7(x))^2}+\frac{1}{((x)-7\frac{1}{x})^2}+\frac{1}{(\frac{1}{x}-7)^2}) $$
And the minimum of this last function is 0.25
