Inequality $\frac{1}{1-abc} + \frac{1}{1-bcd} + \frac{1}{1-cda} + \frac{1}{1-dab} \le \frac{32}{7}$ If $a,b,c,d$ are positive real numbers such that $a^2+b^2+c^2+d^2 = 1$,
Prove that: $$\frac{1}{1-abc} + \frac{1}{1-bcd} + \frac{1}{1-cda} + \frac{1}{1-dab} \le \dfrac{32}{7}$$
I saw this problem is very similar to the problem I have got but with different condition on the variables. The problem in the link suggests a power series expansion of the LHS followed by establishing an inequality of the type: $$\sum_{n=0}^{\infty}(bcd)^n+(cda)^n+(dab)^n+(abc)^n$$
and establishing inequality 
$(bcd)^n+(cda)^n+(dab)^n+(abc)^n\ge (K(a^2+b^2+c^2+d^2))^n$ 
for a positive constant $K$. Also I couldn't imitate the solution provided in the link for my problem. Is there a general method for solving these type of problems ?
 A: We make use of the inequality: $\displaystyle \sum\limits_{cyc} abc \le \frac{1}{16}\left(\sum\limits_{cyc} a\right)^3$ several nice proofs are given here.
(The cyclic sum is taken over $a,b,c,d$)
We have: $\displaystyle \sum\limits_{cyc} (abc)^2 \le \frac{1}{16}\left(\sum\limits_{cyc} a^2\right)^3 = \frac{1}{16}$
and $\displaystyle \max\{abc,bcd,cda,dab\} < \left(\frac{a^2+b^2+c^2+d^2}{3}\right)^{3/2} = \frac{1}{3\sqrt{3}}$.
We need a positive constant $c > 0$, such that: $\displaystyle \frac{1}{1-x} \le  \frac{8}{7} + c\left(x^2 - \frac{1}{64}\right)$
for $x \in \left(0,\frac{1}{3\sqrt{3}}\right)$ atleast.
Since, $\displaystyle \frac{8}{7} + c\left(x^2 - \frac{1}{64}\right) - \frac{1}{1-x} = \left(x - \frac{1}{8}\right)\left(c\left(x+\frac{1}{8}\right) - \frac{8}{7(1-x)}\right)$
Then, $\displaystyle x \le \frac{1}{8} \implies c \le \frac{64}{7(1-x)(1+8x)} = g(x)$ and the minima of $g(x)$ is attained at the point $x = \frac{7}{16}$ in te interval $(0,1)$ and is monotone decreasing the interval $\left(0,\frac{7}{16}\right)$. Thus, we may take $c = g(1/8) = \dfrac{256}{49}$.
Then, we see that $\displaystyle \frac{8}{7} + \dfrac{256}{49}\left(x^2 - \frac{1}{64}\right) \ge \frac{1}{1-x}$ for $x \in \left(0,\frac{3}{4}\right)$.
Thus, $\displaystyle \sum\limits_{cyc} \frac{1}{1-abc} \le \frac{32}{7} + \dfrac{256}{49}\sum\limits_{cyc}\left((abc)^2 - \frac{1}{64}\right) \le \frac{32}{7}$
