<s>[INST] Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$ I have to prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3 $$
is always true for real numbers $a, b, c>0$ with $abc=1$. 
Using the AM-GM inequality I got as far as $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq \frac{b}{\sqrt[a+1]{b}}+\frac{c}{\sqrt[b+1]{c}}+\frac{a}{\sqrt[c+1]{a}}$$
but I do not yet know how to finish my proof from there (if this is helpful at all?!).
 [/INST] Hint: $\frac{1+ab}{1+a}=\frac{abc+ab}{1+a}=ab\big(\frac{1+c}{1+a}\big)$. So what is $\big(\frac{1+ab}{1+a}\big)\big(\frac{1+bc}{1+b}\big)\big(\frac{1+ca}{1+c}\big)$?
</s>