<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] Now I see. Using Especially Lime's comment, we can see that $$\left(\frac{1+ab}{1+a}\right)\left(\frac{1+bc}{1+b}\right)\left(\frac{1+ca}{1+c}\right)=1.$$
Hence, we can set $$x:=\frac{1+ab}{1+a}, y:=\frac{1+bc}{1+b}$$ and our inequality becomes $$x+y+\frac{1}{xy}\geq 3$$ which is equivalent to $$\frac{x+y+\frac{1}{xy}}{3}\geq 1.$$ But using the AM-GM inequality for three variables, we can see that $$\frac{x+y+\frac{1}{xy}}{3}\geq \sqrt[3]{xy\cdot\frac{1}{xy}}=1,$$and our claim follows.
</s>