Hint: First prove that f(-x)=-f(x) holds for x which is not 1 or -1, and for any 0<x,y<1, f(x)<f(y) iff x>y. Then prove $f(\frac{1}{n+4})-f(\frac{1}{n+3})=f(\frac{1}{n^2+7n+11})$ The details and other things need to finish...Hmmm, I leave it for you.

Let $y=-x$, we have: $f(\frac{1}{x})+f(\frac{1}{-x})=f(0)=0$ hence: $-f(t)=f(-t)$ for all $t \in (-1, 0) \cup (0, 1)$. Let $1<x_1 < x_2$, then $f(\frac{1}{x_1})-f(\frac{1}{x_2})=f(\frac{x_1-x_2}{1-x_1x_2}) < 0$, so $f(x)$ is decreasing in $(0,1)$. Then use telescoping, we have $\sum f(\frac{1}{n^2+7n+11})=f(\frac{1}{n+4})-f(\frac{1}{4})$ Since $f(\frac14)+f(\frac12)=f(\frac{2+4}{1+8})=f(\frac23)$ and $\frac{1}{n+4} < \frac{2}{3}$, so $f(\frac{1}{n+4})>f(\frac14)+f(\frac12) \Longrightarrow f(\frac{1}{n+4})-f(\frac14)>f(\frac12)$

shobber wrote: $f(\frac{1}{x_1})-f(\frac{1}{x_2})=f(\frac{x_1-x_2}{1-x_1x_2}) < 0$, Why is this?

indybar wrote: shobber wrote: $f(\frac{1}{x_1})-f(\frac{1}{x_2})=f(\frac{x_1-x_2}{1-x_1x_2}) < 0$, Why is this? This is from condition (2) in the question as well as the fact that $-f(x)=f(-x)$