Korea winter program test: $\min_{v}\deg(v)\le1+2\left(\frac{e}{2}\right)^{0.4}$ for a 3d graph with edges of unit length 
A graph $G$ is given with vertices in 3d space. It has $e$ edges and every pair of vertices are connected iff the distance between them is $1$. It is known that there exist a Hamiltonian cycle in $G$. Prove that for $e>1$,
$$\min_{v}\deg(v)\le1+2\left(\frac{e}{2}\right)^{0.4}\text.$$

I tried some double-counting, and most of the things didn't work out well. Any ideas?
Source: 2018 Korea Winter Mop Practice Test #8
 A: The conditions $ e > 1 $ and existence of a Hamiltonian cycle are in fact irrelevant, and the result holds for any $ G $ with vertices in $ \mathbb R ^ 3 $ and edges as mentioned in the statement of the problem, as long as $ G $ has a finite (nonzero) number of vertices. Let $ n $ be the number of vertices of $ G $, and for any positive integer $ i $ with $ 1 \le i \le n $, let $ d _ i $ be the degree of the $ i $-th vertex of $ G $ in some fixed labeling of vertices. Also, denote $ \min _ { 1 \le i \le n } d _ i $ by $ \delta $. In case $ G $ has an isolated vertex, we have $ \delta = 0 $, and the desired inequality holds. If $ n = 1 $, the only vertex of $ G $ has to be isolated, and we're done. From now on, assume that we have $ n \ge 2 $ and $ d _ i \ge 1 $ for any $ i $ with $ 1 \le i \le n $. For simplifying notations, define $ \bar d = \frac 1 n \sum _ { i = 1 } ^ n d _ i $. Note that we have $ n \bar d = 2 e $. Also, note that $ \delta \le \bar d $, and therefore it suffices to prove $ \bar d \le 1 + 2 \left ( \frac e 2 \right ) ^ { 0.4 } $.
The key observation is that $ G $ cannot have $ K _ { 3 , 3 } $ as a subgraph, which is a consequence of the geometric properties of $ \mathbb R ^ 3 $. That means for any $ 3 $ pairwise distinct vertices $ u $, $ v $ and $ w $ of $ G $, there are at most $ 2 $ vertices of $ G $ that can be taken as the internal node of a claw subgraph of $ G $ whose leaves are $ u $, $ v $ and $ w $. Thus, if $ N $ is the number of claws that are a subgraph of $ G $, we must have
$$ N \le 2 { n \choose 3 } \text . \tag 0 \label 0 $$
On the other hand, each claw subgraph is determined by a vertex of $ G $ together with $ 3 $ of its neighbors, and hence
$$ N = \sum _ { i = 1 } ^ n { d _ i \choose 3 } \text . \tag 1 \label 1 $$
Note that the function $ f : [ 1 , + \infty ) \to \mathbb R $ given by
$$ f ( x ) = { x \choose 3 } = \frac { x ( x - 1 ) ( x - 2 ) } 6 $$
is convex on its domain. Since for any $ i $ with $ 1 \le i \le n $, $ d _ i $ is in the domain of $ f $, by Jensen's inequality we get
$$ f \left ( \frac 1 n \sum _ { i = 1 } ^ n d _ i \right ) \le \frac 1 n \sum _ { i = 1 } ^ n f ( d _ i ) \text , $$
which by \eqref{1}, is equivalent to
$$ f \left ( \bar d \right ) \le \frac N n \text . \tag 2 \label 2 $$
Combining $ n = \frac { 2 e } { \bar d } $ with \eqref{0} and \eqref{2} we have
$$ f \left ( \bar d \right ) \le \frac { \bar d } e f \left ( \frac { 2 e } { \bar d } \right ) \text , $$
which can simply be rewritten as
$$ \bar d ^ 5 - 3 \bar d ^ 4 + 2 \bar d - 4 \bar d ^ 2 + 12 e \bar d \le 8 e ^ 2 \text , $$
or
$$ \bar d ^ 5 - 3 \bar d ^ 4 + 2 \bar d + 2 ( 3 n - 2 ) \bar d ^ 2 \le 8 e ^ 2 \text . \tag 3 \label 3 $$
As $ n \ge 2 $, if we show that
$$ \left ( \bar d - 1 \right ) ^ 5 \le \bar d ^ 5 - 3 \bar d ^ 4 + 2 \bar d + 8 \bar d ^ 2 \text , \tag 4 \label 4 $$
then \eqref{3} gives $ \left ( \bar d - 1 \right ) ^ 5 \le 8 e ^ 2 $, or equivalently $ \bar d \le 1 + 2 \left ( \frac e 2 \right ) ^ { 0.4 } $, which is what was desired. But \eqref{4} can be rewritten as
$$ 2 \bar d ^ 4 - 8 \bar d ^ 3 + 18 \bar d ^ 2 - 5 \bar d + 1 \ge 0 \text , $$
which holds, no matter the value of $ \bar d $.
