Prove that $\frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca} \geq \frac{3}{2}.$ 
Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1}{a+ab}+\dfrac{1}{b+bc}+\dfrac{1}{c+ca} \geq \dfrac{3}{2}.$$ 

I thought about substituting in $abc = 1$ to get $$\dfrac{1}{a+\dfrac{1}{c}}+\dfrac{1}{b+\dfrac{1}{a}}+\dfrac{1}{c+\dfrac{1}{b}} = \dfrac{c}{ac+1}+\dfrac{a}{ab+1}+\dfrac{b}{bc+1}.$$ Then I am not sure what inequality to apply.

Let $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
$$\sum_{\text{cyc}}\frac{1}{\frac{x}{y}+\frac{x}{y}\frac{y}{z}}=\sum_{\text{cyc}}\frac{yz}{xz+xy}$$
Let $yz=c$, $xz=d$, $xy=e$. Then your inequality follows from Nesbitt's inequality.
