Prove that, if deg(v) ≥ (n−2)/3 for every vertex v in G, then G contains at most two connected components
Assume there are are at least 3 connected components. Then there must be a connected component of size $\le\frac{n}{3}$. Let $v$ be a vertex in that connected component. By assumption we have $deg(v)\ge\frac{n-2}{3}$. Since the size of that connected component is at most $\frac{n}{3}$ than it must be that $deg(v)\le\frac{n}{3}-1=\frac{n-3}{3}$. So we get both $deg(v)\ge\frac{n-2}{3}$ and $deg(v)\le\frac{n-3}{3}$, a contradiction. Thus, there must be at most 2 connected components.