USAMO 1983 - Inequalities

Maybe I'm getting better. Maybe I picked out an easy one. Maybe the aura from visiting MIT today hasn't rubbed off yet (happiest place ever btw). Whatever the case, I managed to do an old USAMO problem today, with the help of ACoPS's compilation of obscure inequalities:



We want information about the roots, so it'll help to use our basic polynomial rules to rewrite a and b in terms of the roots. Let these roots be F, G, H, J, K:



Sorry if that last image is hard to read at the end. Try doing view image or something like that.

From now on, we'll simply be interpreting the inequality with these substitutions made. Also, I'm going to use that cyclic sum symbol consistently, because it's much easier than writing it all out.

We need to show that if the inequality is satisfied, then the roots which are represented in it cannot all be real. So our strategy will be to assume the inequality is satisfied as a given, and that all the roots are real, and attempt to find a contradiction. When we find one, then it will imply that one of our assumptions was false. Since we're operating under the pretense that our inequality is true, then our assumption about all the roots being real will be false.

With substitutions made, our inequality is:



Here it helps to know how to square expressions in the form (a + b + c + ....) quickly.

By the way, looking at that, I realize that it's a little hard to follow because it's cramped / ugly. So I recommend writing it out yourself.

Now we can start using well known (well, maybe not well known...) inequalities to find a contradiction. We know from Chebyshev's Inequliaty (WTF! Yes, I'll comment on the obscureness of this later) that:




For sequences of real numbers {a_n} and {b_n} that are "monotonic in the same direction". Or in other words, increasing or decreasing.

This works well for our roots, as since we're assuming they're real, we can, WLOG, arrange them in increasing order and apply Chebyshev's. Let {a_n} be our roots, and let {b_n} ALSO be our roots. Then we get the inequality (after multiplying both sides by 25, which I'll skip right over):





If you line this up with our assumed inequality, then it DIRECTLY contradicts it. Thus, we have our contradiction, and our proof is done.

It's possible to do this with AM-GM too, I believe. And I'm not 100% sure that my solution works, as I have nobody to check it with. However, I found the Chebyshev's solution first, and there doesn't appear to be anything wrong with it. Perhaps I'll outline that method later.

As for the obscureness of Chebyshev's...the only reason I used it was that I was using ACoPS while I was doing the problem, and I tried it and it worked. You don't need to know lots of random inequalities to do good on the USAMO (at least, I hope not). However, you should at least know the basics: AM-GM, the general Mean Inequality, and Cauchy-Schwarz inequality. It might help to know Chebyshev's, but get the basics down first.