Let $G$ be a graph on $>1$ vertices. Recall that its maximum average degree is defined to be
$$M(G) = \max\left\{\frac{2\left|E\_H\right|}{|V\_H|} \colon H \subseteq G, |V\_H| > 1\right\},$$
and its arboricity is equal to
$$A(G) = \max\left\{\left\lceil\frac{|E\_H|}{|V\_H|-1}\right\rceil \colon H \subseteq G, |V\_H| > 1\right\}.$$


In [https://arxiv.org/pdf/1802.15267.pdf, near the end of page 2] it is stated that "it is not difficult to show that $2A(G) - 2 \le \lceil M(G)\rceil$". However, I am not sure what is the reasoning.



Arboricity and average degree
Denote $A(G)=A$. Let $H$ be the subgraph for which $\lceil \frac{|E\_H|}{|V\_H|-1}\rceil=A$. Write $e$ for $|E\_H|$, $v$ for $|V\_H|$. We have $e/(v-1)>A-1$, thus $e>(v-1)(A-1)$. On the other hand, $e\leqslant v(v-1)/2$. Therefore $v/2>A-1$, $v>2A-2$, $v\geqslant 2A-1$. Then $$M(G)\geqslant \frac{2e}v=\frac{2e}{v-1}\cdot \frac{v-1}v>2(A-1)\cdot \frac{2A-2}{2A-1}>2A-3,$$
thus $\lceil M(G)\rceil \geqslant 2A-2$.


