## IMO 2011 Problem 3

Let be a real-valued function defined on the set of real numbers that satisfies

for all real numbers and . Prove that for all .

*IMO 2011 Problem 3*

**Solution:** First, suppose there exists with . This means that . Taking we get . For some , take . Then . For we have . Contradiction. Therefore, for all we have .

For we get . For we get . Assume that . Then it follows by the inequality above that and furthermore .

From the hypothesys . Contradiction. Therefore there is a value with . Then . This proves that .

It is easy now to see that , which means that , and we are done.

For more proofs and informations see the discussion in the following link: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=729&t=418798&&start=20

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Categories: Algebra, Problem Solving
functional equation, IMO

When and you claimed , so . What about ? Is that possible and cancel out ?

Well, when , , in turn, .

Ok, you’ve figured it out. 🙂

This was an amazingly elegant solution.

Doesn’t this assume a special value for f(0) though? that it can’t be infinite?

Why it assumes a special value for ? Of course can’t be infinite, since .

Thanks! Doy.

Thank you for this very elegant solution. I learnt alot here.

I’m glad to hear that 🙂