The proof of that particular case goes like this: Pick one of the six users. Split the other 5 users into friends and non-friends of the chosen user. There wil

The proof of that particular case goes like this: Pick one of the six users. Split the other 5 users into friends and non-friends of the chosen user. There will either be at least 3 friends or at least 3 non-friends of the chosen user. If you can pick 3 friends of the chosen user then if any two of them are friends they plus the original user form a trio of mutual friends. If none of the 3 are friends then you have a trio in which none are friends. Similarly, if you can pick 3 non-friends of the original user then either two of them are non friends and you have a trio of non friends or all three are friends, forming a trio of mutual friends. It's easy to prove but I don't think I'd quite call it "bleedin' obvious". Ramsey's theorem [1], of course, is more general than that and isn't at all obvious (although it's still not very hard to prove). [1] https://en.m.wikipedia.org/wiki/Ramsey%27s_theorem https://en.m.wikipedia.org/wiki/Ramsey%27s_theorem