I need to show that
$$1+\frac{1}{2}d(G) < \max_{H \subseteq G} 1+\delta(H)$$
, where $d(G)$ denotes the average degree of the graph $G$, i.e.
$$d(G) = \frac{1}{|(V)|}\cdot\sum_{v \in V}d(v)$$
and $\delta(G)$ denotes the minimum degree of $G$. I recognize that the right side of the inequality is an upper bound for the chromatic number $\chi(G)$, but I do not understand how I could use this here. Could you help me?
The strict inequality fails if $G$ has no edges, since in that case $$\frac12d(G)=0=\max_{H\subseteq G}\delta(G).$$ Discarding that trivial case, we assume that $G=(V,E)$ where $|V|=n$ and $\max_{H\subseteq G}\delta(H)=k\gt0$.
Choose a vertex $v_1$ of degree $f_1\le k$ in $G$; next choose a vertex $v_2$ of degree $f_2\le k$ in $G-v_1$; next choose a vertex $v_3$ of degree $f_3\le k$ in $G-v_1-v_2$; and so on. In this way we list the vertices of $G$ as $V=\{v_1,\dots,v_n\}$ where $v_i$ has degree $f_i\le k$ in the subgraph induced by $\{v_i,\dots,v_n\}$. Since $f_n=0$, we have $$|E|=\sum_{i=1}^nf_i=\sum_{i=1}^{n-1}f_i\le(n-1)k\lt nk,$$ whence $$\max_{H\subseteq G}\delta(H)=k\gt\frac{|E|}n=\frac12d(G).$$