Prove that for every three positive numbers $a, b, c$:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{a+b+c}{a+b+c+\sqrt[3]{abc}}$$
I tried using $\sum_{\mathrm{cyc}}$ but I haven't got far. I also tried:
$$\sqrt[3]{abc}=\frac{1}{3}\sqrt[3]{3a \cdot 3b \cdot 3c}$$
and according to the HM-GM inequality we can replace the root by:
$$\frac13 \frac{3}{\frac1{3a} +\frac1{3b}+\frac1{3c}}=\frac{1}{\frac13\left(\frac1a+\frac1b+\frac1c\right)}=\frac{3}{ \frac1a+\frac1b+\frac1c}$$
And that's it.
Can you give me a hint or a solution for the question?