IMO Problem 3: Let a function satisfy the relation . Prove that for , .

Let . Then we have . Cancelling on both sides, we get .

For , if , then the above gives a contradiction, as but .

Now we see that . This is because from the above equation, we have . This is possible only if .

Hence so far, we have that for , . Now we shall prove that for all , we have . This will give us the desired contradiction.

We have . Let . Then we have , or . This, coupled with the fact that for we have , we get that for all , we have .

### Like this:

Like Loading...

*Related*

I think the last paragraph is problematic: you have an upper bound of $yf(x)$ from the given condition, not a lower bound.

Sorry if the writing has been confusing, but I’m not getting any bound for . I’m only obtaining a lower bound of for . Coupling that with an upper bound of for for , I get the desired result.

By “an upper bound of $yf(x)$”, I meant the inequality in the first sentence $yf(x) \ge f(f(x)) – f(x+y)$, which is in the counter-direction to the given condition that induces $yf(x) “\le ” f(f(x)) – f(x+y)$.

Ah now I see. I seem to have made a mistake there. I’ll try and fix it. Thanks for bringing my attention to it!