IMO 2011, Problem 3

-March 6, 2020Uncategorized

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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4 thoughts on “IMO 2011, Problem 3”

Hhan Minki says:

March 11, 2020 at 8:18 am

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

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ayushkhaitan3437 says:

March 11, 2020 at 12:07 pm

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.

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1. Hhan Minki says:
  
  March 16, 2020 at 3:22 am
  
  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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  1. ayushkhaitan3437 says:
    
    March 16, 2020 at 4:00 am
    
    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!