[proofplan]
Fix an optimal proper colouring $c : V(G) \to \{1, \ldots, \chi(G)\}$. The preimages $C_i = c^{-1}(i)$ partition $V(G)$ into $\chi(G)$ independent sets, each of size at most $\alpha(G)$. Summing the sizes bounds $|G|$ by $\chi(G)\alpha(G)$, and rearranging yields the stated inequality.
[/proofplan]
[step:Fix an optimal proper colouring and extract its colour classes]
Let $k := \chi(G)$ denote the [chromatic number](/theorems/???) of $G$. By definition of $\chi(G)$, there exists a [proper colouring](/theorems/???)
\begin{align*}
c : V(G) &\to \{1, \ldots, k\} \\
v &\mapsto c(v)
\end{align*}
such that $c(u) \neq c(v)$ whenever $uv \in E(G)$. For each $i \in \{1, \ldots, k\}$ define the $i$-th colour class
\begin{align*}
C_i := c^{-1}(i) = \{v \in V(G) : c(v) = i\}.
\end{align*}
The sets $C_1, \ldots, C_k$ form a partition of $V(G)$: every vertex has exactly one colour, so the $C_i$ are pairwise disjoint and their union is $V(G)$.
[guided]
The hypothesis we extract from "$\chi(G) = k$" is the existence of a proper colouring using $k$ colours — that is, a map $c : V(G) \to \{1, \ldots, k\}$ assigning colours to vertices so that adjacent vertices receive different colours. We want to leverage the adjacency constraint on $c$ to relate $k$ to the independence number $\alpha(G)$, so we look at the structure $c$ imposes on $V(G)$.
Form the colour classes $C_i := c^{-1}(i) = \{v \in V(G) : c(v) = i\}$ for $i = 1, \ldots, k$. These are the fibres of $c$. Since $c$ is a function, every vertex lies in exactly one fibre, so the $C_i$ partition $V(G)$:
\begin{align*}
V(G) = C_1 \sqcup C_2 \sqcup \cdots \sqcup C_k.
\end{align*}
The partition structure is what will let us sum over colour classes in the next step.
[/guided]
[/step]
[step:Observe that each colour class is an independent set]
We claim each $C_i$ is an [independent set](/theorems/???) in $G$. Suppose for contradiction that some $C_i$ contained an edge $uv \in E(G)$ with $u, v \in C_i$. Then $c(u) = i = c(v)$, contradicting that $c$ is proper. Hence $C_i$ contains no edges, and by definition of the [independence number](/theorems/???),
\begin{align*}
|C_i| \leq \alpha(G) \quad \text{for all } i \in \{1, \ldots, k\}.
\end{align*}
[/step]
[step:Sum over colour classes and rearrange]
Since $\{C_1, \ldots, C_k\}$ partitions $V(G)$:
\begin{align*}
|G| = |V(G)| = \sum_{i=1}^{k} |C_i| \leq \sum_{i=1}^{k} \alpha(G) = k \cdot \alpha(G) = \chi(G) \cdot \alpha(G).
\end{align*}
The graph has at least one vertex (else $\chi(G) = 0$ and the inequality is vacuous), and $\alpha(G) \geq 1$ for any non-empty graph, so we may divide by $\alpha(G)$ to obtain
\begin{align*}
\chi(G) \geq \frac{|G|}{\alpha(G)},
\end{align*}
which is the desired inequality.
[guided]
We now combine the two observations. The partition identity gives
\begin{align*}
|G| = |V(G)| = \sum_{i=1}^{k} |C_i|,
\end{align*}
and bounding each summand by $\alpha(G)$ (the maximum size of any independent set) yields
\begin{align*}
|G| \leq \sum_{i=1}^{k} \alpha(G) = k \, \alpha(G) = \chi(G) \, \alpha(G).
\end{align*}
To rearrange into the stated form, we divide by $\alpha(G)$. This division is legitimate provided $\alpha(G) > 0$. If $V(G) = \varnothing$ then $|G| = 0$ and $\chi(G) = 0$, and the inequality $0 \geq 0/\alpha(G)$ is interpreted as vacuous; otherwise any single vertex forms an independent set, so $\alpha(G) \geq 1 > 0$. Dividing gives
\begin{align*}
\chi(G) \geq \frac{|G|}{\alpha(G)},
\end{align*}
as required. The inequality is sharp whenever the colour classes of some optimal colouring all have size exactly $\alpha(G)$, for instance in complete multipartite graphs with equal-sized parts.
[/guided]
[/step]
