Let $a,b,c$ be positive real numbers such that $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$ Prove that $abc \le \frac{1}{8}$.

Question

Let $a,b,c$ be positive real numbers such that $$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$$ Prove that $abc \le \frac{1}{8}$.

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So I tried this problem and came up with a solution, but I'm afraid it might have logical fallacies as I'm not too good playing with inequalities.$$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$$ $\Rightarrow \frac{a}{1+a}-1+\frac{b}{1+b}-1+\frac{c}{1+c}-1=-2$ $\Rightarrow \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$. Now by AM GM inequality, $\begin{aligned} & \frac{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}{3} \geqslant \sqrt[3]{\frac{1}{(1+a)(1+b)(1+c)}} \\ \Rightarrow \quad &\left(\frac{2}{3}\right)^3 \geqslant \frac{1}{(1+a)(1+b)(1+c)} \\ \Rightarrow \quad &(1+a)(1+b)(1+c) \geqslant \frac{27}{8} \end{aligned}$. So the minimum value of the expression $(1+a)(1+b)(1+c)$ is $27/8$. From the original equation, applying am gm inequality, we get,$\frac{\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}}{3} \geqslant \sqrt[3]{\frac{a b c}{(1+a)(1+b)(1+c)}}$ $\Rightarrow \quad \frac{1}{27} \geqslant \frac{a b c}{(1+a)(1+b)(1+c)}$ $\Rightarrow \frac{(1+a)(1+b)(1+c)}{27} \geqslant a b c$. The minimum value of $\frac{(1+a)(1+b)(1+c)}{27}$ is attained when $(1+a)(1+b)(1+c)$ is minimum, which is $\frac {27}{8}$. So $min(\frac{(1+a)(1+b)(1+c)}{27})= \frac {1}{8}$. Hence, $\frac {1}{8} \geq abc$. Is this correct? I want to clarify I'm not looking for any solution but rather a clarification on my solution. Thank you.

Answer 1

The key step in your proof is the following: $$f(a,b,c)\geq M \text{ and } f(a,b,c)\geq g(a,b,c)\implies g(a,b,c)\leq M.$$

This is a flawed reasoning - for example take $f = 2+a^2, g = 1+a^2$ and $M = 1.$ If you are wondering your line of reasoning can be salvaged, then unfortunately no because for $a = b = 1, c = \dfrac{4}{23}$ satisfies: $$(1+a)(1+b)(1+c) = 27abc$$ but $abc = \dfrac{4}{23} > \dfrac{1}{8}.$

Answer 2

Your solution is wrong because the last reasoning is wrong:

From $$\frac{(1+a)(1+b)(1+c)}{27}\geq abc$$ not follows that $abc\leq\frac{1}{8}.$

The right solution: by AM-GM $$\prod_{cyc}\frac{1}{a+1}=\prod_{cyc}\left(\frac{b}{b+1}+\frac{c}{c+1}\right)\geq8\prod_{cyc}\sqrt{\frac{bc}{(b+1)(c+1)}}=\frac{8abc}{\prod\limits_{cyc}(a+1)}.$$