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.

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    $\begingroup$ Let $S_1=\sum 1/a,S_2=\sum 1/(a+b)$, then you can use the Arithmetic-Harmonic Mean to show $S_1>4S_2/(N-1),S_3>16S_4/[3(N-3)]$, so it is true for $N=4$ and $N=5$ $\endgroup$ – Empy2 Jun 27 '15 at 8:51
  • $\begingroup$ I wonder if the other question does imply this one. Surely one could rescale the $a_i$ high enough so that rounding them to the nearest integer doesn't affect the inequality, then model those integers with disjoint subsets $A_i$? $\endgroup$ – Erick Wong Jun 29 '15 at 16:18
  • $\begingroup$ @ErickWong Quite possible, and that's what I thought too. I should have said "doesn't directly imply". $\endgroup$ – nan Jun 30 '15 at 0:09


  • $n$ 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.

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  • $\begingroup$ Your solution always nice! +1 $\endgroup$ – math110 Jun 29 '15 at 16:25

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