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Given $p, q, r \in \mathbb{N}$, let $R(p, q)$ denote the smallest natural number "$n$" such that for any coloring of edges of an $n$-clique by 2 colors : blue or red, there exists either a "blue $p$-clique" or a "red $q$-clique".

Let $R(p, q, r)$ denote the smallest natural number "$n$" such that for any coloring of edges of an $n$-clique by 3 colors : blue, red or yellow, there exists either a "blue $p$-clique", "red $p$-clique" or a "yellow $q$-clique".

Prove that for all integers $p, q \geq 2$, $R(p, q, 3) \leq R(p, q) + R(p-1, q) + R(p, q-1) - 1$.

Please give me some suggestions.

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    Is there a typo in the question? Are $R(p-1,q)$ and $R(p,q-1)$ supposed to be $R(p-1,q,3)$ and $R(p,q-1,3)?$ The inequality you wrote is false e.g. when $p=q=3.$ – 2017-01-15
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    I've already checked, this is the exact problem statement, no typo. – 2017-01-15
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    The inequality is false when $p = q = 3 $ as you said. Why did you think it should be $R(p-1,q,3) $ and $R(p,q-1,3)$ ? – 2017-01-15
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    I also get $17 \leq 11$. – 2017-01-15

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You are familiar with the proof of $$R(p,q)\ \le\ R(p-1,q)+R(p,q-1)\tag1$$ which can also be written in the form $$R(p,q)-1\ \le\ 1+[R(p-1,q)-1]+[R(p,q-1)-1]\tag2$$ which is a good way to think about it, seeing as $R(p,q)-1$ is the largest $n$ for which an $n$-clique can be colored blue & red without making a blue $p$-clique or a red $q$-clique. The inequality (2) can be generalized (with essentially the same proof) to any number of colors; e.g., for three colors $$R(p,q,r)-1\ \le\ 1+[R(p-1,q,r)-1]+[R(p,q-1,r)-1]+[R(p,q,r-1)-1]\tag3$$ (see proof below), which simplifies to $$R(p,q,r)\ \le\ R(p-1,q,r)+R(p,q-1,r)+R(p,q,r-1)-1.\tag4$$ Setting $r=3$ and noting that $R(p,q,2)=R(p,q)$ we get $$R(p,q,3)\ \le\ R(p-1,q,3)+R(p,q-1,3)+R(p,q)-1\tag5$$ which somewhat resembles your inequality $$R(p,q,3)\ \le\ R(p-1,q)+R(p,q-1)+R(p,q)-1.\tag{6}$$ But (6) is incorrect.

Example. For $p=2,q=3,$ the inequality (6) says that $$R(2,3,3)\ \le\ R(1,3)+R(2,2)+R(2,3)-1=1+2+3-1=5$$ but in fact $R(2,3,3)=R(3,3)=6.$

Example. For $p=q=3$ the inequality (6) says that $$R(3,3,3)\ \le\ R(2,3)+R(3,2)+R(3,3)-1=3+3+6-1=11$$ but in fact $R(3,3,3)=17=R(2,3,3)+R(3,2,3)+R(3,3,2)-1.$

Looks like there was a typo in the assignment.

Proof of (3). Let $n=R(p,q,r)-1.$ Color the edges of an $n$-clique with three colors (blue, red, yellow) so that there is no blue $p$-clique, no red $q$-clique, and no yellow $r$-clique. Choose a vertex $v$ and suppose that $v$ is joined to $n_1$ vertices by blue edges, to $n_2$ vertices by red edges, and to $n_3$ vertices by yellow edges. Then $R(p,q,r)-1=n=1+n_1+n_2+n_3$
$\le\ 1+[R(p-1,q,r)-1]+[R(p,q-1,r)-1]+[R(p,q,r-1)-1].$

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    Thank you very much. There is one point I get stuck. Will you please show me how the equation (3) is derived ? – 2017-01-15
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    Sorry, I don't understand why $n_1 \le\ [R(p-1,q,r)-1], n_2 \le\ [R(p,q-1,r)-1], $ and $ n_3 \le\ [R(p,q,r-1)-1]$. – 2017-01-15
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    The clique consisting of the $n_3$ vertices joined to $v$ by yellow edges contains no blue $p$-clique (because the $n$-clique contains no blue $p$-clique), no red $q$-clique (because the $n$-clique contains no blue $q$-clique), and no yellow $(r-1)$ clique (because the $n$-clique contains no yellow $r$-clique). By the definition of $R(p,q,r-1)$ it follows that $n_3\lt R(p,q,r-1),$ and so $n_3\le R(p,q,r-1)-1.$ Similar arguments show that $n_1\le R(p-1,q,r)-1$ and $n_2\le R(p,q-1,r)-1.$ – 2017-01-15
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    All are clear now. Thank you so much for your thorough explanations. – 2017-01-15