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
.
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!