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Let $a$, $b$ and $c$ be positive real numbers such that $$\frac a{1 + b} + \frac b{1 + c} + \frac c{1 + a} = 1.$$ Prove that $$abc \leq \frac 18.$$

I have tried to simply and get answer of this inequalities

Please help me to solve this question.

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closed as off-topic by Prasun Biswas, Martin R, Dave, B. Goddard, Xander Henderson Oct 6 '17 at 15:42

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  • 1
    $\begingroup$ I have tried to simply and get answer of this inequalities $\endgroup$ – Dwip Dalal Oct 6 '17 at 13:02
  • $\begingroup$ @DwipDalal You should always mention what you've tried, no matter how trivial, in the question, so that we can help you better. Also, people will be more willing to help if you show that you have put effort into it. As a personal comment, saying that "I've tried to get the answer" doesn't seem very specific. $\endgroup$ – user472341 Oct 6 '17 at 13:03
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expanding the condition we get $$a^2c+ab^2+bc^2+a^2+b^2+c^2=abc+1$$ and then we have $$1+abc\geq 3\sqrt[3]{(abc)^3}+3\sqrt[3]{(abc)^2}$$ this is equivalent to $$1-2abc\geq 3\sqrt[3]{(abc)^2}$$ expanding we get $$-8(abc)^3-15(abc)^2-6abc+1\geq 0$$ this equivalent to $$(1-8abc)(abc+1)^2\geq 0$$ therefore $$abc\le \frac{1}{8}$$

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  • $\begingroup$ Sonnhard I think it's exactly my solution. I think it's not fair, which you did. $\endgroup$ – Michael Rozenberg Oct 6 '17 at 14:41
  • $\begingroup$ sorry Michael, i have not copied your solution, why do you think so about me? $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '17 at 14:49
  • $\begingroup$ it is not so rairly that two Solutions in mathematics are similiar $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '17 at 14:52
  • $\begingroup$ You posted this solution later. If you see that someone posted the same solution you need to delete your. Otherwise it's not fair. Our posts are not similar. They are the same. $\endgroup$ – Michael Rozenberg Oct 6 '17 at 14:53
  • $\begingroup$ ok if you think so about me, i will delete my solution. $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '17 at 15:00
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Let $abc=w^3$.

Thus, since the expanding gives $$1+abc=\sum_{cyc}(a^2+a^2c),$$ by AM-GM we obtain: $$1+w^3\geq3\sqrt[3]{a^2b^2c^2}+3abc=3w^2+3w^3$$ or $$2w^3+3w^2-1\leq0$$ or $$2w^3-w^2+4w^2-2w+2w-1\leq0$$ or $$(2w-1)(w+1)^2\leq0$$ or $$w\leq\frac{1}{2}$$ or $$abc\leq\frac{1}{8}.$$ Done!

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  • $\begingroup$ You have a $+4a^2$ there when you mean $+4w^2$. $\endgroup$ – Thomas Andrews Oct 6 '17 at 13:31
  • $\begingroup$ It was typo. Thank you Thomas! $\endgroup$ – Michael Rozenberg Oct 6 '17 at 13:35
  • $\begingroup$ Can you please explain me how did you shift from 1st step to 2nd step that is from 1+abc=∑cyc(a2+a2c) to 1+w^3>3a2b2c2√3+3abc $\endgroup$ – Dwip Dalal Oct 7 '17 at 9:18
  • $\begingroup$ @Dwip Dalal Yes, of course! $\sum\limits_{cyc}a^2=a^2+b^2+c^2\geq3\sqrt[3]{a^2b^2c^2}=3\sqrt[3]{w^6}=3w^2$ and $\sum\limits_{cyc}a^2c=a^2c+b^2a+c^2b\geq3\sqrt[3]{a^3b^3c^3}=3\sqrt[3]{w^9}=3w^3$. Because $abc=w^3$. $\endgroup$ – Michael Rozenberg Oct 7 '17 at 9:21

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