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I need to proof this inequality by AM-GM method.
Any ideas how to do it?

$$\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$$

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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 $(p+q+r)^2 \ge 3(pq+qr+rp)$

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  • $\begingroup$ How do you use Cauchy Schwartz in the third step? This is not of the form $|x||y|\ge |x\cdot y|$ right? (are you using the sum norm on $\mathbb{R}^{n}$? $\endgroup$ – Bombyx mori Sep 15 '12 at 15:36
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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!

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