<s>[INST] For $a$, $b$, $c$ the sides of a triangle, show $\sum_{cyc}\frac{\sqrt{bc}\,bc}{a(a+b+c)(b+c-a)}\geq1$ 
Let $a,b,c$ be the side lengths of a triangle. How to prove the following inequality?
  $$\frac{\sqrt{bc}\,bc}{a(a+b+c)(b+c-a)}+\frac{\sqrt{ac}\,ac}{b(a+b+c)(a+c-b)}+\frac{\sqrt{ab}\,ab}{c(a+b+c)(a+b-c)}\geq1$$

Via some numerical experiment and graphing techniques, I am sure it is true. Could we use AM-GM?
My goal is to find an analytic/conventional proof of this inequality. I have tried AM-GM, but it didn't work. Since it is a homogeneous function with degree 0, WLOG, let $a=1.$ Then I plotted the LHS of this inequality using the Maple command "plot3d" which clearly shows that the LHS ≥ 1. 
 [/INST] We have to prove that (see @g.kov's result)
$$3\sqrt[3]{\frac{(abc)^2}{(a+b+c)^3(a+b-c)(b+c-a)(c+a-b)}} \ge 1.$$
By using Ravi Substitution $a = x + y, b = y+z, c = z + x$ for $x, y, z > 0$, the inequality becomes
$$3\sqrt[3]{\frac{[(x+y)(y+z)(z+x)]^2}{64(x+y+z)^3xyz}} \ge 1.$$
Let $p = x+y+z, q = xy + yz + zx, r = xyz$. The inequality becomes
$$3\sqrt[3]{\frac{(pq - r)^2}{64p^3 r}} \ge 1.$$
Since $pq \ge 9r$ and $q^2 \ge 3pr$, we have
$$3\sqrt[3]{\frac{(pq - r)^2}{64p^3 r}} \ge 3\sqrt[3]{\frac{(pq - \frac{pq}{9})^2}{64p^2q^2/3}} = 1.$$
We are done.
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