If $a,b,c$ are positive real numbers, then $\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$ [closed]

If $a$, $b$, and $c$ are positive real numbers such that $abc=1$, then prove that $$\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$$

Progress

I think the relevant concept would be the application of AM/GM inequality.

closed as off-topic by DeepSea, Aditya Hase, user98602, Claude Leibovici, Joonas IlmavirtaDec 7 '14 at 10:20

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• I think the concept would be like the application of AM>=GM. – Pratyush Dec 7 '14 at 5:30
• Some parentheses for denominators would improve the clarity of your problem. An edit has been proposed. If it is not correct, please advise so it can be rolled back or made right. – hardmath Dec 7 '14 at 5:34
• what should be the improvement. I didn't get it.. – Pratyush Dec 7 '14 at 5:38
• since the $a,b, \text{and}\ c$ are moving in a cyclic order couldn't you focus on one piece and use a cyclic summation (Don't know if that's the correct terminology.) $\sum_{a,b,c}$ – Fmonkey2001 Dec 7 '14 at 5:46

Writing $a=x/y$, $b=y/z$, and $c=z/x$, we simplify the LHS: $$\frac{xz}{xy+yz}+\frac{xy}{xz+yz}+\frac{yz}{xy+xz}=\frac{r}{s+t}+\frac{s}{r+t}+\frac{t}{s+r}$$ where $r=xz$, $s=xy$, and $t=yz$. It remains to invoke Nesbitt's inequality.
The linked wikipedia article contains a standard Cauchy-Schwarz proof: $$[(s+t)+(r+t)+(s+r)]\left[\frac{1}{s+t}+\frac{1}{r+t}+\frac{1}{s+r}\right]\geq9$$ which after the simplification of the LHS implies $$\frac{r}{s+t}+\frac{s}{r+t}+\frac{t}{s+r}\geq\frac{3}{2}.$$