Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$

Question

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?!).

Answer 1

Hint: $\frac{1+ab}{1+a}=\frac{abc+ab}{1+a}=ab\big(\frac{1+c}{1+a}\big)$. So what is $\big(\frac{1+ab}{1+a}\big)\big(\frac{1+bc}{1+b}\big)\big(\frac{1+ca}{1+c}\big)$?

Answer 2

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.

Answer 3

By AM-GM$$\sum_{cyc}\frac{1+ab}{1+a}\geq3\sqrt[3]{\prod_{cyc}\frac{1+ab}{1+a}}=3\sqrt[3]{\prod_{cyc}\frac{1+\frac{1}{c}}{1+a}}=$$ $$=3\sqrt[3]{\prod_{cyc}\frac{1+c}{1+a}}=3.$$