Problem: (1999 USAMO – #4) Let () be real numbers such that
Prove that .
Solution: To simply things, let . We wish to show that there exists an .
Our conditions become
Assume for the sake of contradiction that for all and let . We have
But we have , so
However, since and , we have (looking at the parabola it is easy to see; has zeros at ). Therefore
But we also had
so that gives us a contradiction. Hence our assumption must be false and there exists an as desired. QED.
Comment: This wasn't a particularly difficult inequality, but it has some key ideas. Using all parts of the question is important (in this case is actually relevant). Another note is that this didn’t really require any of the classical inequalities, just algebraic manipulation. Lastly, the crucial step was setting converting the real to positive which makes things a whole lot nicer.
Practice Problem: Go take the 2006 Mock AIME 5 below.
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