TITLE: Are at most $1/3$ vertices "kings"?
QUESTION [12 upvotes]: If $G=(V,E)$ is a finite, simple, undirected graph, and $v\in V$, we set $N(v) = \{w\in V:\{v,w\}\in E\}$, and $\text{deg}(v)= |N(v)|$. We say a vertex $v\in V$ is a king if $\text{deg}(v) > \text{deg}(w)$ for all $w\in N(v)$.
In the graph $G=(\{0,1,2\}, \big\{\{0,1\}, \{1,2\}\big\})$, one of the $3$ vertices is a king. Let $\text{King}(G)$ be the set of king vertices.
Question. Is it true that for any finite connected graph $G=(V,E)$ with $|V|>1$ we have $|\text{King}(G)|/|V|\leq 1/3$? If not, how large can this value get?

REPLY [5 votes]: I wanted to find a proof that uses Hall's marriage theorem [1] instead of double-counting.
Given a graph $G=(V,E)$, let $K$ be the set of kings in $V$, and $R:=V \setminus K$ the rest.
Claim: $|K| < |R|$.
Proof Let $G=(V,E)$ be a counterexample that minimizes the sum $|V|+|E|$, so $|K| \ge |R|$. Then $G$ is bipartite, since any edge between nodes in $R$ can be removed.  If $|K|>|R|$ then removing one king would yield a smaller counterexample, so $|K|=|R|$.  If there was a  subset $S$ of $K$  where its neighborhood satisfies $|N(S)|< |S|$, then the induced graph on  $S \cup N(S)$ would be a smaller counterexample.    Thus the Hall condition is met in $G$. Removing from   $G$ a perfect matching of $K$ to $R$  yields a smaller counterexample.
[1] https://en.wikipedia.org/wiki/Hall%27s_marriage_theorem