Let $M$ be a non-empty and finite set, $S_1,...,S_k$ subsets of $M$, satisfying:
(1) $|S_i|\leq 3,i=1,2,...,k$
(2) Any element of $M$ is an element of at least $4$ sets among $S_1,....,S_k$.
Show that one can select $[\frac{3k}{7}] $ sets from $S_1,...,S_k$ such that their union is $M$.
Partial solution: I can find a family of ${13\over 25}k$ such sets that no element in $M$ is in more then 3 set from that family. Thus we have a family of the size ${13\over 25}k$ instead of ${4\over 7}k$.
Say $|M| =n$.
Let' s take any set independently with a probability $p$. Let's mark with $X$ a number of a chosen sets and with $Y$ a number of elements that are ''bad'' i.e. elements which are in at least 4 sets among a chosen sets. Note that $4n\leq 3k$. Then we have $$E(X-Y)=E(X)-E(Y) \geq kp-np^4 \geq kp (1-3p^3/4)$$Since a function $x \mapsto x(1-3x^3/4)$ achives a maximum at $x=\sqrt[3]{1/3}$ we have $E(X-Y)\geq {\sqrt[3]{9}\over 4}k> {13\over 25}k$.
So with the method of alteration we find constant ${\sqrt[3]{9}\over 4}$ which is about $0,051$ worse then ${4\over 7}$.
The question is cross-posted to mathoverflow
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