IMO 2011, Problem 3

IMO Problem 3: Let a function f(x):\Bbb{R}\to \Bbb{R} satisfy the relation f(x+y)\leq -yf(x)+f(f(x)). Prove that for x\leq 0, f(x)=0.

Let x+y=f(x). Then we have f(f(x))\leq f(x)(x-f(x))+f(f(x)). Cancelling f(f(x)) on both sides, we get f(x)(x-f(x))\geq 0.

For x\leq 0, if f(x)>0, then the above gives a contradiction, as f(x)>0 but x-f(x)<0.

Now we see that f(0)=0. This is because from the above equation, we have f(0)(-f(0))\geq 0. This is possible only if f(0)=0.

Hence so far, we have that for x<0, f(x)\leq 0. Now we shall prove that for all y\in\Bbb{R}, we have f(y)\geq 0. This will give us the desired contradiction.

We have yf(x)\geq f(f(x))-f(x+y). Let x=0. Then we have -f(y)\leq 0, or f(y)\geq 0. This, coupled with the fact that for y<0 we have f(y)\leq 0, we get that for all y\leq 0, we have f(y)=0.

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Graduate student

4 thoughts on “IMO 2011, Problem 3

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

  2. Sorry if the writing has been confusing, but I’m not getting any bound for yf(x). I’m only obtaining a lower bound of 0 for f(y). Coupling that with an upper bound of 0 for f(x) for x<0, I get the desired result.

    1. 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 $yf(x) “\le ” f(f(x)) – f(x+y)$.

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