Given positive reals $a,b$ and $c$, such that $$\sum_{cyc}\frac{a}{a+1}=1$$ Show that $$abc\le \frac{1}{8}$$
This problem is fairly easy we can just clear denominators and then use AM-GM. But since I want to get better at using Jensen's Inequality. I tried to use it here,
Note that $$ \sum_{cyc}\frac{a}{a+1}=\sum_{cyc}\frac{1}{1+\frac{1}{a}}=1$$ Let $x=1/a, y=1/b$ and $c=1/z$. Then consider the function $$f(x)=\frac{1}{1+e^x} \quad x>0$$ $f$ is convex and by Jensen we have $$\frac{1}{3}\sum_{cyc}\frac{1}{1+e^x}\ge \frac{1}{1+e^{x+y+z/3}}$$ Substuting $x\to \log x$ we get $$\frac{1}{3}\sum_{cyc}\frac{1}{1+x}=\frac{1}{3}\ge \frac{1}{1+\sqrt[3]{xyz}}$$ Hence $$xyz\ge 8\implies abc\le \frac{1}{8}.$$
Correct? I'm afraid of the substitutions that I used.
Edit:
@MartinR pointed out that the substitution $x\to \log x$ isn't correct since $f$ is convex for $x>0$ but $\log x$ can be negative. How can we fix this?