If the sum of cubes of $a,b,c,d$ is $1$, then $\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$ 
$a,b,c,d>0$ satisfying $a^3+b^3+c^3+d^3=1$. Prove
  $$\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$$

I tried to go the normal way, by Cauchy-Schwarz, but that doesn't work. So I tried to incorporate this newly learned idea, since $a,b,c,d<1$ we can write the left as a power series:
$$\sum_{n=0}^{\infty}(bcd)^n+(cda)^n+(dab)^n+(abc)^n$$
If we can show, $(bcd)^n+(cda)^n+(dab)^n+(abc)^n\ge (K(a^3+b^3+c^3+d^3))^n$ for some suitable constant $K$ we can finish. But I can't really do it. Can someone help me?  
 A: The AM-GM inequality yields
$$ abcd \le \frac{1}{4 \sqrt[3]{4}} \iff abc \le \frac{1}{4 \sqrt[3]{4}d} \iff \frac{1}{1-abc} \le \frac{1}{1 - \frac{1}{4 \sqrt[3]{4} d}}. $$
It follows that 
$$ \sum_{cyclic} \frac{1}{1-abc} \le \sum_{cyclic} \frac{4 \sqrt[3]{4} d}{4 \sqrt[3]{4} d -1}.$$
Therefore it is sufficient to show that
$$ \sum_{cyclic} \frac{4 \sqrt[3]{4} d}{4 \sqrt[3]{4} d -1} \le \frac{16}{3} \iff 4 - \sum_{cyclic} \frac{-1}{4 \sqrt[3]{4} d - 1} \le \frac{16}{3} \iff \sum_{cyclic} \frac{-1}{4 \sqrt[3]{4} d - 1} \ge \frac{-4}{3} \iff \sum_{cyclic} \frac{1}{1 - 4 \sqrt[3]{4}d} \ge \frac{-4}{3}. $$
Now using a form of the Cauchy-Schwarz inequality usually referred to as Titu's Lemma (after Titu Andreescu), we find that
$$ 
\sum_{cyclic} \frac{1}{1 - 4 \sqrt[3]{4}d} \ge \frac{(1 + 1 +1 + 1)^2}{\sum_{cyclic} (1 - 4 \sqrt[3]{4}d)} = \frac{16}{4 - 4 \sqrt[3]{4}} (a + b + c + d).
 $$
Hence it suffices to show that 
$$ \frac{16}{4 - 4 \sqrt[3]{4}(a + b + c + d)} \ge \frac{-4}{3} \iff  \frac{1}{1 -  \sqrt[3]{4}(a + b + c + d)} \ge \frac{-1}{3} \iff 3 \ge -1 + \sqrt[3]{4} (a + b + c + d) \iff 4 \ge \sqrt[3]{4} (a + b + c + d) \iff \sqrt[3]{16} \ge a + b + c + d. $$
The last inequality follows from Hölder's inequality since
$$ (a^3 + b^3 + c^3 + d^3)(1 + 1 + 1 + 1)(1 + 1 + 1 + 1) \ge (a + b + c + d)^3 \iff \sqrt[3]{16} \ge a + b + c + d. $$
Therefore the conclusion follows and equality holds for $ a = b = c = d = \frac{1}{\sqrt[3]{4}}.$  
