To prove $abc\leq\frac 18$

Question

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.

Answer 1

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 $$abc\le \frac{1}{8}$$

Answer 2

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!