I solved IMO problem 5 with the "density" style, this is not that precise actual density from the topology course. High School students usually view 'dense' as "always getting somewhere in between". Note that (until this post got published) this proof is quite different from the proofs given in Mathlinks.
Solution:
We first prove the following lemma:
Proof:
Given $x, y \in \mathbb{Q}_+$ with $y> x$, since $a> 1$ we can choose $k$ large enough so that $\frac{a^k}{x}-\frac{a^k}{y}=a^k \left(\frac{y-x}{xy}\right)> 1$. Therefore there exists an integer $N$ in the interval $\left(\frac{a^k}{y},\frac{a^k}{x}\right)$. Thus
\[\frac{a^k}{y} < N < \frac{a^k}{x}\]
This means $x < \frac{a^k}{N} < y$. $\blacksquare$.
From condition 3, we have $f(a)=a > 0$, thus $f(1)a=f(1)f\left(a\right) \geq f\left(a\right) = a$. Therefore $f(1) \geq 1$.
Next we will prove that $f(n)\geq n$ for all $n \in \mathbb{N}$.
Obviously, we have $f(x_1+x_2+\cdots+x_n)\geq f(x_1)+f(x_2)+\cdots+f(x_n)$.
Thus $f(n)\geq f(1)n \geq n$.
Then we prove that $f$ is monotonely increasing, for any positive rational numbers $y > x$, then $y=x+ t$ for some positive rational number $t$. Thus $f(y)= f(x+t)\geq f(x)+f(t) > f(x)$.
We have $f(a^k)\leq f(a)^k = a^k$ thus for any integers $m$ and $k$ we have $m f\left(\frac{a^k}{m}\right) \leq f(a^k) \leq a^k$ which implies
\[f\left(\frac{a^k}{m}\right)\leq \frac{a^k}{m} \qquad \text{for any integers $k, m$}\]
We will prove that $f(r)\leq r$ for all $r \in \mathbb{Q}_+$. Suppose that there exists a positive rational number $r$ such that $f(r) > r$, fix a natural number $k$ , then $v:=\frac{a^{k+1}}{f(r)-r}$ is a defined positive number. Thus we can take an integer $m$ greater than $v$.
But, by our lemma, between $r+\frac{a^k}{m}$ and $r+\frac{a^{k+1}}{m}$ there exists a rational number of the form $\frac{a^l}{u}$, hence by the monotonity we have
\[f(r) < f\left(r+\frac{a^k}{m}\right) < f\left(\frac{a^l}{u}\right) \leq \frac{a^l}{u} < r+\frac{a^{k+1}}{m} \]
But then $m < \frac{a^{k+1}}{f(r)-r} =v$ a contradiction. Thus for all positive rational numbers $r$ we have $f(r)\leq r$ , in particular $f(n)\leq n$ for all natural numbers, since we also have $f(n)\geq n$ we conclude that $f(n)=n$ for all natural numbers $n$.
For any rational numbers $\frac{b}{c}$ where $b$ and $c$ are positive integers we have $f\left(\frac{b}{c}\right) \geq \frac{f(b)}{f(c)}=\frac{b}{c}$, therefore $f(r)\geq r$ for all $r\in \mathbb{Q}_+$, since $f(r)\leq r$ as well, we conclude that $f(r)=r$. It is easy to check that this function satisfies the given conditions.
IMO 2013 Problem 5
Find all function $f: \mathbb{Q}_+\rightarrow \mathbb{R}$ which satisfies:
a) $f(x+y) \geq f(x)+f(y)$ for all $x, y \in \mathbb{Q}_+$.
b) $f(x)f(y) \geq f(xy)$ for all $x, y \in \mathbb{Q}_+$.
c) There exists $a\in \mathbb{Q}_+$ and $a> 1$ such that $f(a)=a$.
Solution:
We first prove the following lemma:
Lemma: The set $A:=\left\{\frac{a^k}{n} \lvert k, n \in \mathbb{N}\right\}$ is "dense" in $\mathbb{Q}_+$. That is for any $x, y \in \mathbb{Q}_+$ there exists $\theta \in A$ such that $x < \theta < y$.
Proof:
Given $x, y \in \mathbb{Q}_+$ with $y> x$, since $a> 1$ we can choose $k$ large enough so that $\frac{a^k}{x}-\frac{a^k}{y}=a^k \left(\frac{y-x}{xy}\right)> 1$. Therefore there exists an integer $N$ in the interval $\left(\frac{a^k}{y},\frac{a^k}{x}\right)$. Thus
\[\frac{a^k}{y} < N < \frac{a^k}{x}\]
This means $x < \frac{a^k}{N} < y$. $\blacksquare$.
Then we prove that $f$ is monotonely increasing, for any positive rational numbers $y > x$, then $y=x+ t$ for some positive rational number $t$. Thus $f(y)= f(x+t)\geq f(x)+f(t) > f(x)$.
We have $f(a^k)\leq f(a)^k = a^k$ thus for any integers $m$ and $k$ we have $m f\left(\frac{a^k}{m}\right) \leq f(a^k) \leq a^k$ which implies
\[f\left(\frac{a^k}{m}\right)\leq \frac{a^k}{m} \qquad \text{for any integers $k, m$}\]
We will prove that $f(r)\leq r$ for all $r \in \mathbb{Q}_+$. Suppose that there exists a positive rational number $r$ such that $f(r) > r$, fix a natural number $k$ , then $v:=\frac{a^{k+1}}{f(r)-r}$ is a defined positive number. Thus we can take an integer $m$ greater than $v$.
But, by our lemma, between $r+\frac{a^k}{m}$ and $r+\frac{a^{k+1}}{m}$ there exists a rational number of the form $\frac{a^l}{u}$, hence by the monotonity we have
\[f(r) < f\left(r+\frac{a^k}{m}\right) < f\left(\frac{a^l}{u}\right) \leq \frac{a^l}{u} < r+\frac{a^{k+1}}{m} \]
But then $m < \frac{a^{k+1}}{f(r)-r} =v$ a contradiction. Thus for all positive rational numbers $r$ we have $f(r)\leq r$ , in particular $f(n)\leq n$ for all natural numbers, since we also have $f(n)\geq n$ we conclude that $f(n)=n$ for all natural numbers $n$.
For any rational numbers $\frac{b}{c}$ where $b$ and $c$ are positive integers we have $f\left(\frac{b}{c}\right) \geq \frac{f(b)}{f(c)}=\frac{b}{c}$, therefore $f(r)\geq r$ for all $r\in \mathbb{Q}_+$, since $f(r)\leq r$ as well, we conclude that $f(r)=r$. It is easy to check that this function satisfies the given conditions.
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