IMO 2013 · Problem 5

Let \(\mathbb{Q}_{>0}\) be the set of positive rational numbers. Let \(f:\mathbb{Q}_{>0}\to\mathbb{R}\) be a function satisfying the following three conditions:

(i) for all \(x,y\in\mathbb{Q}_{>0}\), we have \(f(x)f(y)\ge f(xy)\);

(ii) for all \(x,y\in\mathbb{Q}_{>0}\), we have \(f(x+y)\ge f(x)+f(y)\);

(iii) there exists a rational number \(a>1\) such that \(f(a)=a\).

Prove that \(f(x)=x\) for all \(x\in\mathbb{Q}_{>0}\).

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

Solution

Fix a rational number \(a>1\) satisfying \(f(a)=a\). Applying (i) to \(a\) and \(1\), we obtain \[ a f(1)\ge a, \] so \(f(1)\ge1\).

We first prove that \(f\) is positive everywhere. Repeated application of (ii) gives \[ f(nt)\ge n f(t) \qquad(n\in\mathbb Z_{>0},\ t\in\mathbb Q_{>0}). \] In particular, \(f(n)\ge n f(1)\ge n>0\). By (i), \[ f(n)f(1/n)\ge f(1)\ge1, \] and hence \(f(1/n)>0\). Thus, for any positive integers \(m,n\), \[ f(m/n)\ge m f(1/n)>0. \] Consequently, \(f\) is strictly increasing: if \(u<v\) are positive rationals, then (ii) yields \[ f(v)\ge f(u)+f(v-u)>f(u). \]

Next, we prove that \(f(x)\le x\) for every positive rational \(x\). By (i) and induction, \[ f(a^k)\le a^k \qquad(k\in\mathbb Z_{>0}), \] because \(f(a^{k+1})\le f(a^k)f(a)=a f(a^k)\).

Fix \(x\in\mathbb Q_{>0}\), and for sufficiently large positive integers \(k\), put \[ n_k=\left\lfloor\frac{a^k}{x}\right\rfloor\ge1. \] Since \(n_kx\le a^k<(n_k+1)x\), repeated application of (ii) and monotonicity give \[ n_k f(x)\le f(n_kx)\le f(a^k)\le a^k<(n_k+1)x. \] Therefore \[ f(x)<x+\frac{x}{n_k}. \] As \(a>1\), we have \(n_k\to\infty\), so this implies \(f(x)\le x\).

Finally, for any \(x\in\mathbb Q_{>0}\), positivity, (i), and the upper bound just proved imply \[ 1\le f(1)\le f(x)f(1/x) \le \frac{f(x)}{x}\le1. \] Every inequality must therefore be an equality, and in particular \(f(x)=x\), as required.

The solution was checked against the official IMO shortlist solutions.