<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] (I) By rearrangement inequality:
$\displaystyle \begin{align} &\sum_{cyc} \frac{1}{b(a+b)} \ge \sum\limits_{cyc} \frac{1}{b(a+c)} \tag{1} \\ \iff & \sum_{cyc} \frac{1}{b(a+b)} \ge \sum_{cyc} \frac{1}{2}\left(\frac{1}{b(a+b)} + \frac{1}{b(a+c)}\right) = \sum_{cyc} \frac{1}{2}\left(\frac{1}{b(a+b)} + \frac{1}{c(a+b)}\right) \\ \iff & \sum_{cyc} \frac{1}{b(a+b)} \ge \frac{1}{2}\sum_{cyc} \frac{b+c}{bc(a+b)} \end{align}$
By Am-Gm Inequality :
$$\sum_{cyc} \frac{b+c}{bc(a+b)} \ge 3\sqrt[3]{\prod\limits_{cyc} \frac{b+c}{bc(a+b)}} = 3$$
Thus establishing desired inequality.
Note: $(1)$ can be viewed as a consequence of CS as well.
$$\sum_{cyc} \left(\frac{1}{b(a+b)} - \frac{1}{b(a+c)}\right) \ge 0 \iff \sum_{cyc} \frac{c-b}{b(a+b)(a+c)} \ge 0 \\ \iff \sum_{cyc} \frac{c^2-b^2}{b} \ge 0 \iff \sum_{cyc} \frac{c^2}{b} \ge \sum_{cyc} b$$
(II) Substituting $\displaystyle a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$:
The inequality required to prove becomes:
$$\sum\limits_{cyc} \frac{x^2}{z^2+xy} \ge \frac{3}{2}$$
We may rewrite LHS as $\displaystyle \sum\limits_{cyc} \frac{x^4}{x^2z^2+x^3y}$ and apply Cauchy-Schwarz Inequality:
$$\sum\limits_{cyc} \frac{x^4}{x^2z^2+x^3y} \ge \frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc} x^2z^2 + \sum\limits_{cyc} x^3y}$$
So it suffices to prove that: $$2(x^2+y^2+z^2)^2 \ge 3\sum\limits_{cyc} x^2z^2 + 3\sum\limits_{cyc} x^3y \\ \iff (x^4+y^4+z^4) + \sum\limits_{cyc} (x^4 + x^2y^2) \ge 3\sum\limits_{cyc} x^3y$$
This is the consequence of adding the following:
(i) The Rearrangement Inequality: $x^4+y^4+z^4 \ge x^3y+y^3z+z^3x$
(ii) The Am-Gm Inequality: $\displaystyle \sum\limits_{cyc} (x^4 + x^2y^2) \ge 2\sum\limits_{cyc} \sqrt{x^6y^2} = 2\sum\limits_{cyc} x^3y$
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