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 $f(y)=0$.

4 thoughts on “IMO 2011, Problem 3”

1. Hhan Minki says:

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

2. ayushkhaitan3437 says:

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. Hhan Minki says:

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 $yf(x) “\le ” f(f(x)) – f(x+y)$.

1. ayushkhaitan3437 says:

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!