For positive $a$, $b$, $c$, show $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{a+b+c}{a+b+c+\sqrt[3]{abc}}$ 
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?
 A: I'll prove a stronger inequality:

Let $a$, $b$ and $c$ be positive numbers. Prove that:
  $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}.$$

Indeed, by C-S 
$$\sum_{cyc}\frac{a}{a+b}=\sum_{cyc}\frac{a^2}{a^2+ab}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}.$$
Thus, it's enough to prove that
$$\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ or
$$\frac{a+b+c}{(a+b+c)^2-ab-ac-bc}\geq\frac{1}{a+b+c-\sqrt[3]{abc}}$$ or
$$ab+ac+bc\geq(a+b+c)\sqrt[3]{abc},$$ which says that in the last case our inequality is proven.
Now, let $$ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}.$$ 
Thus, by C-S again we obtain:
$$\sum_{cyc}\frac{a}{a+b}=\sum_{cyc}\frac{a^2c^2}{a^2c^2+c^2ab}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}$$ and it remains to prove that
$$\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ or
$$\frac{(ab+ac+bc)^2}{(ab+ac+bc)^2-(a+b+c)abc}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}},$$ which is
$$ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}$$ exactly.
Done!
