I was showing some basic graph property and though I could formally prove it, I can't wrap my head around the intuition of why this property holds. The property is the following:
Let $G$ be a graph of $m$ edges. Show that for every $d \in \mathbb N$, if $m < \dfrac{d \times (d+1)}{2}$, then there is a node $v$ such that $d(v) < d$, where $d(v)$ is the degree of the node $v$.
My proof is by contradiction using the properties that the degree of every node is bounded by (cardinal of G) - 1 and by the equality $2m = \sum_{v \in G} d(v)$.
Any explanation or visual ideas on why this property is indeed true would by greatly appreciated.
Not sure whether this is what you are looking for. Let $v$ be a vertex of $G$. If $v$ has degree less than $d$, then we are done. If not, then $v$ has at least $d$ neighboring vertices, say, $v_1,v_2,\ldots,v_k$ with $k\geq d$. Now, the total number of edges of the vertex-induced subgraph with vertices $v_1,v_2,\ldots,v_k$ and all their neighbors is at least $$\frac{1}{2}\,\left(k+\sum_{i=1}^k\,\deg(v_i)\right)\,.$$ (Here is probably a possible visual idea. Well, try to imagine what this subgraph look like and why the expression above is a lower bound on the number of edges in this subgraph.)
By the assumption that there are less than $\dfrac{d(d+1)}{2}$ edges in $G$, we get $$\frac{1}{2}\,\left(k+\sum_{i=1}^k\,\deg(v_i)\right)<\frac{d(d+1)}{2}\,.$$ Thus, $$\frac{1}{k}\,\sum_{i=1}^k\,\deg(v_i)<\frac{d(d+1)}{k}-1\leq d\,.$$ By the Pigeonhole Principle, $\deg(v_j)<d$ for some $j\in\{1,2,\ldots,k\}$.