How to show the convexity of this set Is the set, $S=\{\bf x \in \mathbb{R}^n: \sum_{i=1}^{n} \frac{e^{x_i}}{1+e^{x_i}}=1 \}$, a convex set?
 A: It is not a convex set. I have the following counterexample for $n=2$ (calculated by hand, so it might be wrong):
Take $a=(\ln(\frac{1}{4}),\ln(\frac{2}{5}))$,$b=(\ln(\frac{2}{5}), \ln(\frac{1}{4}))$. This gives us $\frac{1}{2}(a+b) = (\frac{1}{2}\ln(\frac{1}{10}),\frac{1}{2}\ln(\frac{1}{10}))$
We have $a,b \in S$ but $\frac{1}{2}(a+b) \notin S$
A: This should help. It is always a good idea to plot. Note : $1$-dimensional convex sets are subsets of lines. You don't expect something having to do with exponentials to be a line segment (a priori it could have been, but your first guess should be no). 
A: (Because the question changed ( $\frac{e^x}{1-e^x}$ became $\frac{e^x}{1+e^x}$), i decided to write another answer)
It is still not convex. Your calculations for $n=2$ are correct, but e.g. for $n=3$ it is not correct, here is another counterexample (again, calculated by hand):
Take $a=(\ln(\frac{1}{2}),\ln(\frac{1}{5}),0)$,$b=(\ln(\frac{1}{5}), \ln(\frac{1}{2}),0)$. This gives us $\frac{1}{2}(a+b) = (\frac{1}{2}\ln(\frac{1}{10}),\frac{1}{2}\ln(\frac{1}{10}),0)$
We have $a,b \in S$ but $\frac{1}{2}(a+b) \notin S$
