Does a cubic graph with $2n$ vertices admit a minimal cover with $n1$ vertices?

A subset of vertices $S$ in a graph $G$ is called a dominating set if every vertex in $G$ is in $S$ or is connected to a vertex in $S$. The size of the smallest dominating set in a graph is called the domination number of $G$. Even though your question asks about a minimum (vertex/edge) covering, what you seem to be interested in is the dominating number of a cubic graph. Not only does the domination number satisfy the bound in your question, but it can be improved further. Bruce Reed showed in "Paths, stars, and the number three", Combin. Probab. Comput. 5 (1996) 277295, that every cubic graph has its domination number bounded by $3V/8$, where $V$ is the number of vertices in your graph. The bound is achieved for some graphs on 8 vertices. This bound has been more recently improved on by Kostochka and Stodolsky. I believe the conjectured best bound is $5V/14$, but it is not known if infinitely many graphs achieve it. 


Pick any vertex. It is adjacent to three other vertices. Now repeatedly pick an edge joining two vertices, call them $x$ and $y$, such that $y$ has not yet been accounted for, and choose $x$. Each such choice adds one more to the vertices accounted for. When you've exhausted the graph, you've picked $n1$ vertices such that each of the other $n+1$ vertices is adjacent to at least one of the $n1$. 


Dear Gjergji Zaimi, thank you for your reply but I think there is something wrong with this value $ 3  V  / $ 8 because the following cubic graph, http://www.dharwadker.org/independent_set/fig2a.gif, has 6 vertices and a dominating set with three vertices. Is not This? 


Looks wrong. The Heawood graph is 3regular on 14 vertices and its vertex cover is 7.


