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Prove for every undirected graph $G$: $$\frac{1}{\chi (G)} + \frac{\beta (G)}{|V(G)|} \leqslant 1$$

where $\beta(G)$ is the size of the minimum vertex cover in $G$, $\chi(G)$ is the chromatic number of $G$, and $V(G)$ is the set of vertices of $G$.

I have no idea how to begin.

Any ideas?

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    I would suggest defining your variables $G, \chi(G), \beta(G), V(G)$ so that someone not familiar with graph theory can understand your question. – 2017-02-25
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    Did it, thanks @Aguila – 2017-02-25

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Let $V' \subseteq V$ be a minimum vertex cover of $G$; that is, each edge in $G$ is incident to at least one vertex in $V'$. It is immediate to know that $V - V'$ is a maximum independent set of $G$ and $$ \frac{|V'|}{|V|} + \frac{|V - V'|}{|V|} = 1 $$ Note that the vertices in $G$ can be partitioned into $\chi(G)$ monochromatic classes of vertices, where each class can contain at most $|V - V'|$ vertices. Therefore, $$ |V- V'| \cdot \chi(G) \geq |V| \Rightarrow \frac{1}{\chi(G)} \leq \frac{|V - V'|}{|V|} $$ We conclude $$ \frac{\beta(G)}{|V|} + \frac{1}{\chi(G)} \leq \frac{|V'|}{|V|} + \frac{|V - V'|}{|V|} = 1 $$