The definition of a line graph is as follows:
Given a graph G$G$, its line graph L(G)$L(G)$ is a graph such that
- each vertex of L(G)$L(G)$ represents an edge of G$G$.
- two vertices of L(G)$L(G)$ are adjacent if and only if their corresponding edges share a common endpoint ("are incident") in G$G$.
I am curious about what kind of graph satisfies $\vert E(L(G))\vert < \vert E(G) \vert$$\lvert E(L(G))\rvert < \lvert E(G) \rvert$.
It is a well-known fact that $$\vert E(L(G)) \vert=\sum_{i=1}^n {d_i \choose 2}$$$$\lvert E(L(G)) \rvert=\sum_{i=1}^n {d_i \choose 2}$$ where the degree sequence of $G$ is $d_1,\cdots,d_n$$d_1,\dotsc,d_n$.
Now I have $$\vert E(L(G))\vert - \vert E(G) \vert<0 \Leftrightarrow \sum_{i=1}^n d_i(d_i-2) < 0$$$$\lvert E(L(G))\rvert - \lvert E(G) \rvert<0 \Leftrightarrow \sum_{i=1}^n d_i(d_i-2) < 0.$$
Obviously, isolated vertices and cycles can be ignored since $d_i(d_i-2)=0$.
So remaining cases are $d_i=1$ or $d_i \geq 3$.
I observed that if $d_j \geq 3$ then $$d_j < \sqrt{n}+1$$$$d_j < \sqrt{n}+1.$$
Otherwise $d_j(d_j-2) \geq n-1$ so that $$\sum_{i=1}^n d_i(d_i-2) \geq d_j(d_j-2)+(-1)\cdot(n-1) \geq 0$$$$\sum_{i=1}^n d_i(d_i-2) \geq d_j(d_j-2)+(-1)\cdot(n-1) \geq 0.$$
So I considered the union of stars with the degree of 'center'‘center’ less than $\sqrt{n}+1$, but it didn't work well.
I have no idea how to proceed from here to find the properties of $G$.
Would you give me an advice?
