Prove that if $\delta(G) \geq 2$ then $G$ contain a cycle
I tried to prove this using proof by contrapositive.
I assume that $G$ has no cycle and show that $\delta(g) <2$.
The smallest cycle we can have is $C_3$. if $G$ has no cycle that mean $G$ has at least one vertex of degree $1$ or $0$. Meaning $\delta (G) <2$.
Is this proof sound acceptable? It look a little bit too short to me.