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 ?

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    $\begingroup$ This inequality is old inequality $\endgroup$ – math110 Dec 26 '14 at 5:09
  • $\begingroup$ @math110 can you give me a few references ? Thank you ! $\endgroup$ – sciona Dec 27 '14 at 0:13

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, $\displaystyle \sum\limits_{cyc} \frac{1}{1-abc} \le \frac{32}{7} + \dfrac{256}{49}\sum\limits_{cyc}\left((abc)^2 - \frac{1}{64}\right) \le \frac{32}{7}$

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    $\begingroup$ It's nice answer!+1,in fact $$\dfrac{8}{7}+\dfrac{256}{49}(x^2-\dfrac{1}{64})-\dfrac{1}{1-x}=-\dfrac{(4x-3)(8x-1)^2}{49(1-x)}\ge 0,0<x<3/4$$ $\endgroup$ – math110 Dec 26 '14 at 5:43
  • $\begingroup$ @math110 That's how I inferred the inequality too ! :-) $\endgroup$ – r9m Dec 26 '14 at 5:48
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    $\begingroup$ can you explain why consider this form term $c(x^2-\dfrac{1}{64})$ with constant$\dfrac{1}{64}$? $\endgroup$ – math110 Dec 26 '14 at 5:51
  • $\begingroup$ @math110 I was initially trying to get a quasilear upper bound of the form $\dfrac{8}{7}+c(x - \dfrac{1}{8})-\dfrac{1}{1-x}$ for a positive constant $c$, but this has a single root at $x = 1/8$ thus we cannot establish the inequality for $x \le \dfrac{1}{8}$. So I replaced $x$ with $x^2$ and made $\frac{1}{8}$ a double root of the equation (like the identity you wrote) to ensure a positive minimum in the interval $(0,1)$. $\endgroup$ – r9m Dec 26 '14 at 6:01
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    $\begingroup$ oh,I konw because this inequality if and only if when $a=b=c=d=\dfrac{1}{2}$,so $abc=\dfrac{1}{8}$,so $1-abc=1-x$ when $x=\dfrac{1}{8}$.Nice Thank you $\endgroup$ – math110 Dec 26 '14 at 6:06

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