TITLE: Inequality with reciprocals of $n$-variable sums
QUESTION [6 upvotes]: Let $a_1,a_2,\ldots,a_n$ be positive real numbers. Is it always true that
$$\sum_{i=1}^n\frac{1}{a_i}-\sum_{1\leq i<j\leq n}\frac{1}{a_i+a_j}+\sum_{1\leq i<j<k\leq n}\frac{1}{a_i+a_j+a_k}-\cdots+\frac{(-1)^{n-1}}{a_1+\ldots+a_n}>0?$$
The inequality is trivially true when $n=1,2$. For $n=3$, we have $$\frac{1}{a_1}>\frac{1}{a_1+a_2},\frac{1}{a_2}>\frac{1}{a_2+a_3},\frac{1}{a_3}>\frac{1}{a_3+a_1}, \frac{1}{a_1+a_2+a_3}>0.$$ For $n=4$ it is harder to compare terms directly.
Note: This is related to this question about inclusion-exclusion-like sum, but neither question implies the other.

REPLY [6 votes]: Let 

$n$ be any fixed positive integer.
$S_k$, $k = 1,\ldots, n$ be the collection of subsets of $\{ 1, 2, \ldots, n \}$ with exactly $k$ elements.

The sum we want can be rewritten as
$$\mathcal{S} \stackrel{def}{=} \sum_{k=1}^n (-1)^{k-1} \sum_{I \in S_k} \frac{1}{\sum_{i\in I} a_i }$$
For any $n$ positive numbers $b_1, b_2,\ldots, b_n$, if we expand out
$\prod_{j = 1}^n ( 1 - b_j )$, we have
$$\prod_{j=1}^n ( 1 - b_j ) 
= 1 - \sum_{k=1}^n (-1)^{k-1} \sum_{I \in S_k} \prod_{i\in I} b_i$$
Substitute $b_j$ by $e^{-a_j t}$ in this expansion and notice
$$\frac{1}{\sum_{i \in I} a_i} 
= \int_0^\infty e^{-(\sum_{i\in I}a_i)t} dt
= \int_0^\infty \left(\prod_{i\in I} e^{-a_i t} \right) dt 
$$
The sum $\mathcal{S}$ we want can be rewritten as an integral
$$\mathcal{S} = \int_0^\infty \left(1 - \prod_{j=1}^n ( 1 - e^{-a_j t} )\right) dt$$
Since the integrand is non-negative and not identical zero, above integral and hence the sum $\mathcal{S}$ we want is positive.