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

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

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.

• Nov 27, 2022 at 20:05
• @MartinR yes there are solutions available. But I wanted to check if my solution is correct or not, as I stated at the end of my post . Nov 27, 2022 at 20:16
• There's a particular tag you should include when that is what you are looking for; it has been added now. Nov 27, 2022 at 20:57

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 $$abc = \dfrac{4}{23} > \dfrac{1}{8}.$$
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 $$\prod_{cyc}\frac{1}{a+1}=\prod_{cyc}\left(\frac{b}{b+1}+\frac{c}{c+1}\right)\geq8\prod_{cyc}\sqrt{\frac{bc}{(b+1)(c+1)}}=\frac{8abc}{\prod\limits_{cyc}(a+1)}.$$