satisfies for all real numbers and . Show that for all .
First, set gives us inequality for all .
Let the inequality be denoted by .
gives us for all . Similarly, we have , by switching and .
Adding the two inequalities up, we get . Let in this inequality, we see that .
If , we see that from , we get either or . We now try to rule out the latter case, by observing that replace in the inequality with and we get . This demands to be non-positive.
Now, since is positive, is non-positive, we have for that certain , which contradicts our initial inequality . Thus this cannot happen and .
It remains to show . Now, we re-use the inequality by letting be any negative number, which gives off and thus .
gives . Assume now that , and let , which requires since is positive, and thus we see for large enough, , which is impossible as now is negative and . Contradiction. Thus has been wrong which combined with gives .
Comment: The functional inequality does not lead to simple non-trivial solution, and we thus hope to exploit the “badly-behavedness” of the condition by reducing it to something simple yet non-trivial. (In this problem’s case, the inequaity .) Since the condition is an inequality, it is not easy to derive an equality out of it. In this solution we established equality from both sides, to thus force the equality to be true.
The motivation for the substitution is to yield a term on one side, being symmetric to the term on the other side. Since on the other side only appears once and appears outside , we can afford to substitute a relatively complicated form of . Once we reach , substituting is simply giving up the strength of that inequality while preserving its non-triviality. It should be noted that substituting leads to weaker inequality .
The difficulty of the problem lies in the difficulty in dealing with the iterated application of , given that it is not easy to find a non-trivial solution to this equation–under the pressure of IMO and the anxiety of the windmill problem. (I personally do not know of any elementary non-trivial solution to the inequality, which nonetheless makes it a great problem.) Also, functional inequality is much less common than functional equations. I reckon that if it appeared on the IMO as Q2 instead with the windmill problem being Q3, it would be better done–people would likely spend more time on it. And it proves itself not as challenging as the windmill problem after all!
This solution is probably very close to the official solution.
The problem is proposed by Igor Voronovich, Belarus, and appeared in IMO 2011 as Q3.