Balti Tee 2025 · Ülesanne 3

We begin by noticing that $f(0)\ge0$, since

$$f(0)+f(0)\ge f(0+0)\quad\Longrightarrow\quad f(0)\ge0.$$

We will use the following lemma.

Lemma. For all $n\in\mathbb Z_{>0}$ and $x\in\mathbb Q$, we have $nf(x)\ge f(nx)$.

Proof. We use induction on $n$. The case $n=1$ is immediate. Assuming $nf(x)\ge f(nx)$, we get

$$\begin{aligned} (n+1)f(x)&=nf(x)+f(x)\\ &\ge f(nx)+f(x)\\ &\ge f(nx+x)\\ &=f((n+1)x), \end{aligned}$$

which completes the induction.

Next, we prove that for all rational $x<0$ and $y>0$,

$$\frac{f(x)}x\le\frac{f(y)}y.$$

Write $x=-a/b$ and $y=c/d$, where $a,b,c,d$ are positive integers. Then

$$\begin{aligned} f(ac)+f(-ac)&\ge f(0)\ge0,\\ f\!\left(\frac cd\,ad\right)+f\!\left(-\frac ab\,bc\right)&\ge0,\\ ad\,f(y)+bc\,f(x)&\ge0. \end{aligned}$$

Dividing by $ac>0$ gives

$$\frac1y f(y)-\frac1x f(x)\ge0,$$

hence $\frac{f(x)}x\le\frac{f(y)}y$.

Now let

$$S=\left\{\frac{f(x)}x:x<0\right\}, \qquad T=\left\{\frac{f(y)}y:y>0\right\}.$$

Since every $p\in S$ and $q\in T$ satisfy $p\le q$, there exists a real number $\alpha$ such that $p\le\alpha\le q$ for all $p\in S$ and $q\in T$. Choose such an $\alpha$.

If $x<0$, then $f(x)/x\le\alpha$ is equivalent to $f(x)\ge\alpha x$. If $x=0$, we have $f(0)\ge0=\alpha\cdot0$. If $x>0$, then $f(x)/x\ge\alpha$ is equivalent to $f(x)\ge\alpha x$. Therefore $f(x)\ge\alpha x$ for all $x\in\mathbb Q$.