Minimum and maximum of sum of inverse degree of a graph
Nice question!

There is no simple closed form of $D$ for general graphs that involves $|V|$ since $D$ can take at least two different values for any connected graph with more than 2 vertices. $D=1-\frac1{|V|}$ for a start graph and $D=\frac{|V|}4 &gt; 1-\frac1{|V|}$ for a cycle graph. In fact, I would believe that the number of different values of $D$ for a graph of $n$ vertices grows exponentially with respect to $n$.

Yuval's tight lower bound
Claim: $D \geq 1−1/n$ for a connect graph with $n$ vertices, where the equality holds for and only for star graphs.
Proof. Let $G$ be a connect graph with $m$ edges. Since $G$ is connected, $m\ge n-1$, where the equality holds for and only for tree graphs. Let $e=\{u,v\}$ be an edge in $G$.
$$\begin{aligned}
d(u)+d(v) &amp;= \Sigma_{s\in E,\, s \text { contains u}} 1 + \Sigma_{s\in E,\, s \text { contains v}} 1 \\
&amp;= (1 + \Sigma_{s\in E,\, s \text { contains u},\, s\neq e} 1) + (1 + \Sigma_{s\in E,\, s \text { contains v},\, s\neq e} 1)  \\
&amp;= 2 + (\Sigma_{s\in E,\, s \text { contains u},\, s\neq e} 1 + \Sigma_{s\in E,\, s \text { contains v},\, s\neq e} 1)  \\
&amp;\le 2 + \Sigma_{s\in E,\, s\neq e} 1\\
&amp;= 2 + (m -1)\\
&amp;= m +1
\end{aligned}$$
The inequality above holds since the only edge that contains both $u$ and $v$ is $e$. The equality $d(u)+d(v)=m+1$ holds when every edge besides $e$ has a common vertex with $e$.
So,
$$
D = \sum_{e=\{u,v\} \in E} \frac{1}{d(u)+d(v)} \ge m\, \frac1{m+1} = 1 - \frac1{m+1} \ge 1-\frac1n 
$$
The equality $D=1-\frac1n$ holds when there are $n-1$ edges in total and every two edge has a common vertex, which means the graph is a star graph.

There is no positive constant $k$ such that $D \geq |V|/k$, since $|V|/k$ goes to infinity while $1-1/|V|$ stays below 1 when $|V|$ goes to infinity.