Since $2 |E_H|\le |V_H|(|V_H|-1)$, we get $|E_H|/(|V_H|-1)\le |E_H|/|V_H|+1/2$, taking ceiling and maximum over $H$ in LHS we get $$A(G)\le \lceil M(G)/2+1/2\rceil\le \lceil M(G)\rceil/2+1.$$

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 $\lceil M(G)\rceil \geqslant 2A-2$.

Since $2 |E_H|\le |V_H|(|V_H|-1)$, we get $|E_H|/(|V_H|-1)\le |E_H|/|V_H|+1/2$, taking ceiling and maximum over $H$ in LHS we get $$A(G)\le \lceil M(G)/2+1/2\rceil\le \lceil M(G)/2\rceil+1.$$

Since $2 |E_H|\le |V_H|(|V_H|-1)$, we get $|E_H|/(|V_H|-1)\le |E_H|/|V_H|+1/2$, taking ceiling and maximum over $H$ in LHS we get $$A(G)\le \lceil M(G)/2+1/2\rceil\le \lceil M(G)\rceil/2+1.$$