Here is an attempt.
It is well known that $a,b,c$ are the sides of a triangle if and only if you can find numbers $x,y,z >0$ so that $a=x+y, b=x+z, c=y+z$. Your inequality becomes then
$\frac{x+y}{2x+4y}+\frac{x+z}{4x+2z}+\frac{y+z}{4z+2y} \geq 1 \,;\, \forall x,y,z >0 \,.$
This inequality reduces after horrible computations to
$ x^2y+y^2z+z^2x \geq 3xyz $
But this is a bad solution.
Here is a better idea, cannot complete the solution though:
The equation is equivalent to
$\sum_{cyc}\frac{x+y}{x+2y} \geq 2$
or
$\sum_{cyc}1-\frac{y}{x+2y} \geq 2$
or
$1 \geq \sum_{cyc}\frac{y}{x+2y} \,.$
Probably the easiest approach from here would be to denote $x+2y=m, y+2z=n, z+2x=p$ and solve for $x,y,z$. This suggest that probably it would had been best to denote $m=3a-b+c, n=3b-c+a, p=3c-a+b$ from beginning.