(Reinhard Diestel Graph Theory, GTM 173, edition 3)
Theorem 1.3.4 (Alon, Hoory, Linial 2002): Let $G$ be a graph. If $d(G) \geq d \geq 2$ and $g(G) \geq g \in \mathbb N$, then $|G| \geq n_0(d,g)$, where $n_0(d,g) : = 1 + d \sum_{i=0}^{r-1} (d-1)^i$ if $g=: 2r+1$ is odd, or $n_0(d,g): = 2 \sum_{i=0}{r-1}(d-1)^i$ if $g=:2r$ is even.
Here, $d(G)$ is the average degree of $G$, and $g(G)$ is the girth of $G$ (the minimum length of a cycle in $G$).
As far as I see, no prove is given on this textbook.
There is an exercise on page 30 (the 6th),
Prove the weakening of Theorem 1.3.4 obtained by replacing average with minimum degree. Deduce that $|G| \geq n_0(d/2,g)$ for every graph $G$ as given the theorem.
I think this is equivalent to
(*) For $c$ the minimum degree of $G$, and $g$ the girth of $G$, $|G| \geq n_0(c,g)$. For $d$ the average degree of $G$, and $g$ the girth of $G$, $|G| \geq n_0(d,g)$.
Is it alright for me to equal the exercise to ()? Would anyone please give me some advice on the proof of the exercise or ()?