# Connection between max independent set and graph coloring

Is there any connection between the size of the largest independent set in a graph, and the minimum number of colors required to color the graph? I know that we can potentially color all the vertices in the largest independent set in the same color, but we know nothing about the rest of the vertices (besides being a vertex cover). Am I wrong?

There are relationships in the form of inequalities involving number of vertices $$n(G)$$ of a graph $$G$$.
Let $$\alpha(G)$$ be the size of a maximal independent set in $$G$$ and $$\chi(G)$$ the chromatic number, that is the smallest number of colors needed to color a graph $$G$$. Then obviously
$$\chi(G) + \alpha(G) \leq n + 1,$$ since we can obtain one coloring with $$n - \alpha(G) + 1$$ colors, if we color all the vertices in the maximum independent set with one color and the remaining vertices with pairwise distinct colors.
In any G it trivially holds: $$n(G) \leq \chi(G) \cdot \alpha(G) \tag{1}$$
and from this we also get $$2 \sqrt{n(G)} \leq \chi(G) + \alpha(G),$$ because we can expand $$0 \leq \left(\sqrt{\chi(G)} - \sqrt{\alpha(G)} \right)^{2} \\ 0 \leq \chi(G) - 2 \sqrt{\chi(G) \cdot \alpha(G)} + \alpha(G)$$ and use (1) to get $$2 \sqrt{n(G)} \leq \chi(G) + \alpha(G).$$ Combining both inequalities we get $$2 \sqrt{n(G)} \leq \chi(G) + \alpha(G) \leq n(G) + 1.$$