<s>[INST] 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.
 [/INST] Notice that the inequality is equivalent to 
$(abc+1)(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)})+3\ge 6$
This is equivalent to proving that 
$\frac{1+abc+a+ab}{a(1+b)}+\frac{1+abc+b+bc}{b(1+c)}+\frac{1+abc+c+ca}{c(1+a)}\ge6$.
However, $\frac{1+abc+a+ab}{a(1+b)}+\frac{1+abc+b+bc}{b(1+c)}+\frac{1+abc+c+ca}{c(1+a)}=\frac{1+a}{a(1+b)}+\frac{1+b}{b(1+c)}+\frac{1+c}{c(1+a)}+\frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1} \ge 6 (\because AM-GM)$.
Therefore, our proof is done. 
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