Problem 5

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}

f(x)f(y)\ge f(xy)

; (ii) for all

x,y\in\mathbb{Q}_{>0}

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}

Step 1 of 6: First bounds from the two inequalities

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f(1)\ge 1, \qquad f(nx)\ge n f(x) \text{ for } n\in\mathbb{Z}_{>0}

Detailed analysis

Plugging $x=1,y=a$ into (i) gives $f(1)f(a)\ge f(a)$ , and since $f(a)=a>0$ this forces $f(1)\ge1$ . Induction on $n$ using (ii) then gives $f(nx)\ge nf(x)$ for every positive integer $n$ and $x\in\mathbb{Q}_{>0}$ ; in particular $f(n)\ge nf(1)\ge n$ .