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The question is :

Let $a,$$b,$$c$ be positive real numbers such that $ \frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+a} =1$ Then find the maximum value of $abc$.

I just blindly simplified the equation and wrote $abc$ on LHS and other terms on RHS so I concluded if I could find the maximum value of RHS that would help but i am unable to do that also .please tell what would be the correct approach and method for the question .This question requires a subjective approach (not trial and error).

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  • $\begingroup$ I see. I'll delete my comment, thank you for clarifying. $\endgroup$
    – Clayton
    Aug 5, 2019 at 15:01
  • $\begingroup$ @AkshajBansal reread my comment. I don't understand the question $\endgroup$
    – mathworker21
    Aug 5, 2019 at 15:05
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    $\begingroup$ @AkshajBansal also, doesn't $a=b=c=1/2$ make each fractional term on the LHS equal to $\frac{1}{3}$? $\endgroup$
    – mathworker21
    Aug 5, 2019 at 15:05
  • $\begingroup$ Ok thanks i edited the question $\endgroup$
    – Akshaj Bansal
    Aug 5, 2019 at 15:10
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    $\begingroup$ @mathworker21 We need to find a maximal value of $abc$ under the condition. $\endgroup$
    – Michael Rozenberg
    Aug 5, 2019 at 15:32

2 Answers 2

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$\sum\limits_{cyc}\frac{a}{1+b}=\frac{\sum\limits_{cyc}a^2c+\sum\limits_{cyc}ac+\sum\limits_{cyc}a^2+\sum\limits_{cyc}a}{\sum\limits_{cyc}a+\sum\limits_{cyc}ac+1+abc}=1$

$\sum\limits_{cyc}a^2c+\sum\limits_{cyc}a^2=abc+1$

$1=\sum\limits_{cyc}a^2c+\sum\limits_{cyc}a^2-abc\geq 3abc+3\sqrt[3]{a^2b^2c^2}-abc\geq 2abc+3\sqrt[3]{a^2b^2c^2}=2t^3+3t^2$

Where $t=\sqrt[3]{abc}$. Can you finish from here?

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After full expanding by AM-GM we obtain: $$1+abc=\sum_{cyc}(a^2+a^2c)\geq3\left(\sqrt[3]{a^2b^2c^2}+abc\right).$$ The equality occurs for $a=b=c.$

Can you end it now?

I got $\frac{1}{8}$ as the answer.

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  • $\begingroup$ Yes thanks i got that too $\endgroup$
    – Akshaj Bansal
    Aug 5, 2019 at 19:17
  • $\begingroup$ @Akshaj Bansal You are welcome! $\endgroup$
    – Michael Rozenberg
    Aug 5, 2019 at 19:20

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