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Let $a,b,c$ be positive reals such that $abc = 1$. Prove that $\sum_{\text{cyc}} \frac{1}{b(a+b)} \ge \frac{3}{2}$.

Well, I tried like this before : Since $abc=1$, the inequality can then be written as $$\sum_{\text{cyc}} \frac{ac}{a+b} \ge \frac{3}{2}$$ But then, I don't know how to manipulate further so that I can use the AM-GM inequality, or maybe others such as Cauchy Schwarz, etc. Can anyone help? That will be really appreciated! Thanks a lot!

Vann
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    What is meant by $\sum_{\text{cyc}}$? Does this mean $\frac{1}{a(b+a)} + \frac{1}{a(c+a)}+\frac{1}{b(a+b)}+\frac{1}{b(c+b)}+\frac{1}{c(a+c)}+\frac{1}{c(b+c)}$? – kccu Jan 09 '20 at 15:32
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    No...that is $\sum_{\text{sym}}$, in this case $\sum_{\text{cyc}} \frac{1}{b(a+b)}$ means $\frac{1}{b(a+b)} + \frac{1}{c(b+c)} + \frac{1}{a(c+a)}$. – Vann Jan 09 '20 at 15:36

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