Finding maximum value

Question

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).

Answer 1

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

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

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

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

Answer 2

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.