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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1$\begingroup$ I have tried to simply and get answer of this inequalities $\endgroup$– Dwip DalalOct 6, 2017 at 13:02
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$\begingroup$ @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. $\endgroup$– user472341Oct 6, 2017 at 13:03
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2 Answers
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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 $$abc\le \frac{1}{8}$$
answered Oct 6, 2017 at 13:25
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$\begingroup$ Sonnhard I think it's exactly my solution. I think it's not fair, which you did. $\endgroup$ Oct 6, 2017 at 14:41
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$\begingroup$ sorry Michael, i have not copied your solution, why do you think so about me? $\endgroup$ Oct 6, 2017 at 14:49
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$\begingroup$ it is not so rairly that two Solutions in mathematics are similiar $\endgroup$ Oct 6, 2017 at 14:52
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$\begingroup$ 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. $\endgroup$ Oct 6, 2017 at 14:53
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$\begingroup$ ok if you think so about me, i will delete my solution. $\endgroup$ Oct 6, 2017 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!
answered Oct 6, 2017 at 13:11
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$\begingroup$ You have a $+4a^2$ there when you mean $+4w^2$. $\endgroup$ Oct 6, 2017 at 13:31
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$\begingroup$ 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 $\endgroup$ Oct 7, 2017 at 9:18
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$\begingroup$ @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 $abc=w^3$. $\endgroup$ Oct 7, 2017 at 9:21