Prove G has at most 2 commponents

Question

Let G be a graph of order n. Prove that if $deg(u)+deg(v)\geq n-2$ for every pair $u,v$ of nonadjacent vertices of G, then G has at most two components.

MY ATTEMP: Assume to the contrary that G has 3 components, thus since all the components are disconnected, we can say that $deg(v_i)\leq\frac{n}{3}-1$ for some natural number $i$. Therefore $deg(v_i)+deg(v_j)\leq \frac{2n}{3}-2$. From here I am stuck I am unsure where to go from here to get a contadiction.

Answer 1

Sure, contradiction works.

Suppose that there are at least three components, pick vertices $u$ and $v$ inside two different connected components, suppose they have $a$ and $b$ vertices each, clearly $a+b

We also have $d(u)\leq a-1$ and $d(v)\leq b-1$, it follows that $d(u)+d(v)\leq a+b-2