8
$\begingroup$

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 ?

$\endgroup$
2
  • 1
    $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
5
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.