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Hello I have question about inequality :

Imagine we have $a,b,c$ real numbers with $1>a>-1,\;1>b>-1,\;1>c>-1$ with the condition $$\left(\frac{1+a}{1-a}\right)^3+\left(\frac{1+b}{1-b}\right)^3+\left(\frac{1+c}{1-c}\right)^3=3$$

What can we say about these differences :

$\dfrac{1}{1-|ab|}-\dfrac{1}{1-a^2}$

$\dfrac{1}{1-|ac|}-\dfrac{1}{1-c^2}$

$\dfrac{1}{1-|bc|}-\dfrac{1}{1-b^2}$

I mean can we have these kind of inequality :

$\dfrac{1}{1-a^2}-\dfrac{1}{1-|ab|}\geq 2\epsilon$

$\dfrac{1}{1-c^2}-\dfrac{1}{1-|ac|}\geq -\epsilon$

$\dfrac{1}{1-b^2}-\dfrac{1}{1-bc}\geq -\epsilon$

Where $\epsilon$ depends of $a,b,c$ ?

Edit : It's related to this

Finally I found my inequality with the condition of the beginning we have :

$$(0.5+|a|)(\dfrac{1}{1-a^2}-\dfrac{1}{1-|ab|})+(0.5+|b|)(\dfrac{1}{1-b^2}-\dfrac{1}{1-|bc|})+(0.5+|c|)(\dfrac{1}{1-c^2}-\dfrac{1}{1-|ac|})\geq 0$$

Thanks a lot .

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  • $\begingroup$ firstly, for this kind of task you need to apply an inequality separating the $a$-, $b$- and $c$-s from the first summands. (something like $\left|ab\right|\leq a^2 + b^2$). secondly, by permutation of variables, w.l.o.g. it is sufficient to investigate the a-b-case only. $\endgroup$ – Max Aug 16 '17 at 16:48
  • $\begingroup$ can you please write the original question? I mean the inequality you are trying to prove.... $\endgroup$ – MAN-MADE Aug 16 '17 at 16:48
  • $\begingroup$ when $a=0$ ,$b=0$, $c=0$ then first inequality gets invalid. So last two inequalities seem to be correct $\endgroup$ – user450210 Aug 16 '17 at 16:50
  • $\begingroup$ For the final inequality I forgot the condition $|a|\geq|b|\geq|c|$. $\endgroup$ – max8128 Aug 17 '17 at 10:54

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