Finding maximum value

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).

• I see. I'll delete my comment, thank you for clarifying. – Clayton Aug 5 '19 at 15:01
• @AkshajBansal reread my comment. I don't understand the question – mathworker21 Aug 5 '19 at 15:05
• @AkshajBansal also, doesn't $a=b=c=1/2$ make each fractional term on the LHS equal to $\frac{1}{3}$? – mathworker21 Aug 5 '19 at 15:05
• Ok thanks i edited the question – Akshaj Bansal Aug 5 '19 at 15:10
• @mathworker21 We need to find a maximal value of $abc$ under the condition. – Michael Rozenberg Aug 5 '19 at 15:32

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

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

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

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

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.

• Yes thanks i got that too – Akshaj Bansal Aug 5 '19 at 19:17
• @Akshaj Bansal You are welcome! – Michael Rozenberg Aug 5 '19 at 19:20