Let $a_1,a_2,\ldots,a_n\geq 1$, and let $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$. Consider the sum

$$S(f)=\sum_{i}f(a_i)-\sum_{i<j}f(a_i+a_j)+\sum_{i<j<k}f(a_i+a_j+a_k)-\cdots+(-1)^{n-1}f(a_1+\cdots+a_n).$$

This question shows that if $f(x)=\frac1x$, then $S(f)>0$ for all $a_1,\ldots,a_n$. If we perturb $f$ a tiny bit, say $f(x)=\frac{1}{x}-\frac{1}{100x^{100}}$, I would imagine that $S(f)>0$ still always holds. But the proof method for $f(x)=\frac1x$ is hard to generalize to other functions. Can we prove it in some other way?

More generally, is there a theorem out there stating sufficient conditions under which $S(f)>0$ always holds?

Convex Functions, Partial Orderings, and Statistical Applications( books.google.de/… ). I am not fully sure, though, since I might be misreading the $\cdots$ in the formula. $\endgroup$ – darij grinberg Jun 30 '15 at 15:21