IMO 2011 · Problem 3

Let \(f:\mathbb{R}\to\mathbb{R}\) be a real-valued function defined on the set of real numbers that satisfies \[ f(x+y)\le yf(x)+f(f(x)) \] for all real numbers \(x\) and \(y\). Prove that \(f(x)=0\) for all \(x\le 0\).

Official text of the problem. Official problem paper (PDF)

Solution

Set \(a=f(0)\) and \(b=f(a)\). Taking \(x=0\) in the given inequality gives \[ f(t)\le at+b\qquad\text{for every }t\in\mathbb R. \] Consequently, taking \(y=-a\) in the original inequality, we obtain \[ f(x-a)\le -af(x)+f(f(x)) \le -af(x)+af(x)+b=b. \] Since \(x-a\) ranges over all real numbers, \[ f(t)\le b\qquad\text{for every }t\in\mathbb R. \] Thus \(b=f(a)\) is the global maximum of \(f\).

The substitution \((x,y)=(a,0)\) now gives \[ b=f(a)\le f(f(a))=f(b)\le b, \] so \(f(b)=b\). Also, taking \((x,y)=(a-1,1)\), we find \[ b=f(a)\le f(a-1)+f(f(a-1))\le 2b, \] and therefore \(b\ge0\).

Using \(f(b)=b\), the substitutions \((x,y)=(b,-b)\) and \((x,y)=(b,a-b)\) give, respectively, \[ a\le -b^2+b, \qquad b\le (a-b)b+b. \] Since \(b\ge0\), these imply \[ 0\le(a-b)b\le -b^3. \] Hence \(b=0\). The identity \(f(b)=b\) then yields \(f(0)=0\), and the global bound above becomes \(f(t)\le0\) for every real \(t\).

Finally, let \(x<0\). Taking \(y=-x\) in the original inequality gives \[ 0=f(0)\le -xf(x)+f(f(x))\le -xf(x). \] Because \(-x>0\), this implies \(f(x)\ge0\). Together with \(f(x)\le0\), it follows that \(f(x)=0\). We already know this at \(x=0\), completing the proof.

The solution was checked against the official IMO shortlist solutions.