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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?!).

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

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    $\begingroup$ this is a very good idea! $\endgroup$ – Dr. Sonnhard Graubner May 6 '17 at 20:09
  • $\begingroup$ Thank you! It's an ingenious idea to first calculate the product of the three fractions instead of the sum - the rest of the solution is quite easy then. $\endgroup$ – mxian May 6 '17 at 20:17
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By AM-GM$$\sum_{cyc}\frac{1+ab}{1+a}\geq3\sqrt[3]{\prod_{cyc}\frac{1+ab}{1+a}}=3\sqrt[3]{\prod_{cyc}\frac{1+\frac{1}{c}}{1+a}}=$$ $$=3\sqrt[3]{\prod_{cyc}\frac{1+c}{1+a}}=3.$$

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  • $\begingroup$ I don't really see why $3\sqrt[3]{\prod_{cyc}\frac{1+\frac{1}{c}}{1+a}}$ should be equal to $3\sqrt[3]{\prod_{cyc}\frac{1+c}{1+a}}$. Could you explain that, please? $\endgroup$ – mxian May 7 '17 at 9:33
  • $\begingroup$ @mxian $\sqrt[3]{\prod\limits_{cyc}\frac{1+\frac{1}{c}}{1+a}}=\sqrt[3]{\prod\limits_{cyc}\frac{1+c}{c(1+a)}}=\sqrt[3]{\prod\limits_{cyc}\frac{1+c}{1+a}}$ because $abc=1$. $\endgroup$ – Michael Rozenberg May 7 '17 at 14:18
  • $\begingroup$ Ok, I see. Thank you! $\endgroup$ – mxian May 7 '17 at 15:51
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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.

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