# Solving Through The Use Of Handshakes.

Let $G$ be a graph. Use the Handshake Theorem to prove that $\delta(G)\nu(G) \le 2\varepsilon(G) \le \Delta(G)\nu(G)$.

So the first step to solve this I know is that you need to know what everything stands for, like

• $\delta(G)$= minimum degree
• $\nu(G)$ number vertices,
• $\varepsilon(G)$ number edges, and
• $\Delta(G)$ = maximum degree $G$.

Then I know we use the formula $d(G) = 2\varepsilon (G)$. But this is where I get lost because I do not know how to apply the theorem. Any tips regarding how to solve this enigma would be kindly appreciated.

Handshake Theorem: $\sum_{v\in V(G)} d(v)=2\varepsilon(G)$.
Hence, $\delta(G)\nu(G)\leq\sum_{v\in V(G)} d(v)=2\varepsilon(G)\leq \Delta(G)\nu(G)$, where $V(G)$ is the vertex set of $G$.