2013 IMO Problems/Problem 5: Difference between revisions

Latest revision as of 00:32, 19 November 2023

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 $x\in\mathbb Q_{>0}$ .

Solution

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@@ Line 8: / Line 8: @@
 Prove that <math>f(x)=x</math> for all <math>x\in\mathbb Q_{>0}</math>.
-Proposed by Bulgaria
+==Solution==
+{{solution}}
+==See Also==
+*[[2013 IMO]]
+{{IMO box|year=2013|num-b=4|num-a=6}}

2013 IMO (Problems) • Resources
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2013 IMO Problems/Problem 5: Difference between revisions

Latest revision as of 00:32, 19 November 2023

Problem

Solution

See Also