<s>[INST] If $abc=1$ then $\sum\limits_{cyc}^{}{\frac{1}{b(a+b)}}\ge \frac{3}{2}$ 
If $abc=1$ for positive $a,b,c$, then $\sum\limits_{cyc}^{}{\dfrac{1}{b(a+b)}}\ge \dfrac{3}{2}$

I have tried the following,in decreasing order of success:
1)AM-GM:$a+b+c\ge 3$ and $ab+bc+ca\ge 3$
2)Substituting $1=abc$ yields nothing
3)Substituting $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ yields something weird
4)Rearrangement inequality on the sequences $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and $\frac{1}{a+b},\frac{1}{b+c},\frac{1}{c+a}$
5)Titu's lemma doesn't help
A little nudge in the right direction would help.
 [/INST] Hint:
Indeed, as @Macavity pointed out, the sum $s(a,b,c)$ only has  circular symmetry. In fact it is not hard to show ( the same method as below) that if $a\le b\le c$ then 
$s(a,b,c)\le s(c,b,a)$. 
Here is a method amenable to a computer algebra system. As suggested, substitute $a=u/v$, $b=v/w$, $c=w/u$. Get an equivalent inequality homogeneous of degree $6$ in $u$,$v$, $w\ge 0$ 
$$2 u^4 v^2-3 u^3 v^3+2 u v^5+2 u^5 w-3 u^4 v w+4 u^2 v^3 w-
3 u v^4 w+4 u^3 v w^2-\\-6 u^2 v^2 w^2+2 v^4 w^2-3 u^3 w^3+4 u v^2 w^3-3 v^3 w^3+2 u^2 w^4-3 u v w^4+2 v w^5\ge 0$$
with circular symmetry. It is enough to consider the cases $u\le v \le w$ and $u \ge v \ge w$.
In the first case, make the substitution $u=p$, $v=p+q$, $w=p+q+r$, with $p$, $q$, $r\ge 0$. One gets
$$8 p^4 q^2+18 p^3 q^3+15 p^2 q^4+6 p q^5+q^6+8 p^4 q r+31 p^3 q^2 r
+38 p^2 q^3 r+22 p q^4 r+\\+5 q^5 r+8 p^4 r^2+41 p^3 q r^2
+66 p^2 q^2 r^2+46 p q^3 r^2+13 q^4 r^2+14 p^3 r^3+43 p^2 q r^3+\\
+43 p q^2 r^3+17 q^3 r^3+9 p^2 r^4+17 p q r^4+10 q^2 r^4+2 p r^5+2 q r^5$$
clearly positive, with equality when $q=r=0$, that is, the $a$,$b$,$c$ are equal.
The other case is treated similarly, giving the positive expression
$$8 p^4 q^2+18 p^3 q^3+15 p^2 q^4+6 p q^5+q^6+8 p^4 q r+23 p^3 q^2 r+22 p^2 q^3 r+8 p q^4 r+q^5 r+8 p^4 r^2+33 p^3 q r^2+42 p^2 q^2 r^2+18 p q^3 r^2+3 q^4 r^2+14 p^3 r^3+35 p^2 q r^3+23 p q^2 r^3+5 q^3 r^3+9 p^2 r^4+11 p q r^4+2 q^2 r^4+2 p r^5$$
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