Two inequalities with inclusion-exclusion Let $S$ be a base set of size $n$.
Also denote $k$ of $S$'s subsets by $A_1,...,A_k\in 2^S$ such that $\bigcup_i A_i=S$.
Furthermore, let $n_\alpha=\lvert\bigcap_{i\in\alpha}A_i\rvert$ be the size of the intersection of sets in the index set $\alpha\subset I=\{1,...,k\}$.
For example, $n_{1,3,4}=\lvert A_1\cap A_3\cap A_4\rvert$ if $k\ge4$.
We know that $\forall i\in I: n_i\ge3$ and $\sum_i n_i>\sum_{\alpha\subset I} (-1)^{\lvert\alpha\rvert+1}{n_\alpha\choose2}$.
If I could show that $\sum_{\alpha\subset I} (-1)^{\lvert\alpha\rvert+1}n_\alpha\le2k$, that would help me a lot with my project.
I have tried an inductive proof but got stuck.
I have also tried to represent the task with two lattices (or semi-lattices, I am not sure) and find some mapping there but that failed too.
I have checked Richard Stanley's book on Enumerative Combinatorics but I did not see anything that would connect inclusion-exclusion sums of numbers and binomial coefficients.
I would be very grateful for any ideas.
Thank you.
 A: For $r=0,1,\dots,k$ let $U_{r}$ denote the set of elements that have exactly $r$ occurrences in the $A_{i}$; so actually: $$x\in U_r\iff\sum_{i=1}^k1_{A_i}(x)=r$$
Then:
$$1_{U_{r}}=\sum_{\alpha\subseteq I}\binom{\left|\alpha\right|}{r}\left(-1\right)^{\left|\alpha\right|-r}1_{\bigcap_{i\in\alpha}A_{i}}\tag1$$
and consequently: $$\left|U_{r}\right|=\sum_{\alpha\subseteq I}\binom{\left|\alpha\right|}{r}\left(-1\right)^{\left|\alpha\right|-r}n_{\alpha}\tag2$$
This under the convention that a binomial coefficient $\binom{m}s$ takes value $0$ if $s\notin\{0,\dots,m\}$.
In your case we have $S=\bigcup_{i=1}^kA_i$ so that every element has an occurrence
in at least one of the $A_{i}$.
That means that $U_{0}=\varnothing$ and consequently:
$$0=\left|U_{0}\right|=\sum_{\alpha\subseteq I}\binom{\left|\alpha\right|}{0}\left(-1\right)^{\left|\alpha\right|}n_{\alpha}=\sum_{\alpha\subseteq I}\left(-1\right)^{\left|\alpha\right|}n_{\alpha}$$

If you are interested in a proof of $(1)$ then please let me know.
