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I have to conduct two proofs for graph coloring, but I don't have idea where to start. I need to show the following: $G $ is a graph and K is complement of $G$,

$1)\; \chi (G) + \chi(K) \leq |V| + 1$,

$2)\; \chi(G) \cdot \chi(k) \leq ((|V| + 1)^2)/4$.

Could you help me conduct these proofs, please?

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    Prove (1) by induction on the number of vertices. (2) follows from (1) by [the inequality of the geometric and arithmetic means](https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means), $\sqrt{ab}\le\frac{a+b}2.$ – 2017-01-18
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    Yeea, I see that I need to somehow involve number of vertices, because of the right side of inequality, but I don't know where to start actually, I mean I don't know how to properly connects it with chromatic number. – 2017-01-18
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    Okey, I found how to get second with your help, thanks and found first in the another question. Thanks a lot :) – 2017-01-18

2 Answers 2

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For $2)$: $$ (|V|+1)^2 \ge (\chi(G)+\chi(K))^2 = (\chi(G)-\chi(K))^2+4\chi(G)\chi(K) \ge 4\chi(G)\chi(K) $$

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I found for the first question answer here: the Nordhaus-Gaddum problems for chromatic number of graph and its complement

According to @bof comment, the second one I can derive from the first one and sqrt(ab) <= (a+b)/2 inequality. we know that X(G) and X(K) are positive so we can square it and we got:

ab <= (a^2 + b^2 + 2ab)/4

going throught we get

x^2 + y^2 >= 2xy.

Using first inequality we square it and we get

X(G)^2 + X(K)^2 + 2X(G)X(K) <= (|V|+1)^2.

Combine this two inequalites give us

4X(G)X(K) <= (|V| + 1)^2

what we exactly wanted to prove.