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2013 IMO Problems/Problem 5: Difference between revisions

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==Problem==
Let <math>\mathbb Q_{>0}</math> be the set of all positive rational numbers. Let <math>f:\mathbb Q_{>0}\to\mathbb R</math> be a function satisfying the following three conditions:
Let <math>\mathbb Q_{>0}</math> be the set of all positive rational numbers. Let <math>f:\mathbb Q_{>0}\to\mathbb R</math> be a function satisfying the following three conditions:


Revision as of 11:49, 21 June 2018

Problem

Let $\mathbb Q_{>0}$ be the set of all 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 $f(x)f(y)\geq f(xy)$; (ii) for all $x,y\in\mathbb Q_{>0}$, we have $f(x+y)\geq f(x)+f(y)$; (iii) there exists a rational number $a>1$ such that $f(a)=a$.

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

Proposed by Bulgaria

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