Prove $\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$ using AM-GM I need to proof this inequality by AM-GM method.
Any ideas how to do it?  
$$\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$$
 A: We may assume without loss of generality that $abc = k^3$, which enables us to make the substitution $\displaystyle a = \frac{kq}{p}, b = \frac{kr}{q}, c = \frac{kp}{r}$.
Now, $\displaystyle a(1+b) = \frac{kq}{p} \left(1+ \frac{kr}{q} \right) = \frac{k(q+kr)}{p}$.
Thus, the inequality reduces to proving (after cancelling $k$ from the denominator on both sides) $$\frac{p}{q+kr} +\frac{q}{r+kp} +\frac{r}{p+kq} \ge \frac{3}{1+k} $$
By Cauchy Schwarz, we get $$\left(\frac{p}{q+kr} +\frac{q}{r+kp} +\frac{r}{p+kq} \right) ( p(q+kr) + q(r+kp) + r(p+kq)) \ge (p+q+r)^2$$
Thus, we have that $$\frac{p}{q+kr} +\frac{q}{r+kp} +\frac{r}{p+kq} \ge \frac{(p+q+r)^2}{(1+k)(pq+qr+rp)} \ge \frac 3{1+k}$$ due to the well known $(p+q+r)^2 \ge 3(pq+qr+rp)$
A: By AM-GM we obtain:
$$(1+abc)\sum_{cyc}\frac{1}{a(1+b)}+3=\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}=$$
$$=\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{a(1+b)}\geq\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc},$$
which says that
$$\sum_{cyc}\frac{1}{a(1+b)}\geq\frac{\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}-3}{1+abc}=\frac{3\left(1-\sqrt[3]{abc}+\sqrt[3]{a^2b^2c^2}\right)}{(1+abc)\sqrt[3]{abc}}=\frac{1}{(1+\sqrt[3]{abc})\sqrt[3]{abc}}.$$
Done!
