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.

Look at $\frac{1}{b(a+b)}$ and $\frac{1}{a(a+b)}$. Adding together yields $\frac{1}{ab}$. So the sum is just 
$$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} = \frac{a + b + c}{abc}$$
The denominator is 1 and the numerator is at least 3 by AM-GM for positive $a,b,c$.
