Thailand Mathematical Olympiad

Suppose that there are $a, b \in \mathbb{R}$ such that $f(a + b) \neq f(a) + f(b)$. By the assumption we have $f(a + b) = -f(a) - f(b)$. If $f(a + b) = 0$ then $f(a) = -f(b)$.
$$\begin{align*} |f(2a + 2b)| &= |f(a + (a + b + b))| \\ &= |f(a) + f((a + b) + b)| \\ &= |f(a) + f(a + b) + f(b)| \text{ or } |f(a) - f(a + b) - f(b)| \\ &= 0 \text{ or } |f(a) - (-f(a) - f(b)) - f(b)| \\ &= 0 \text{ or } 2|f(a)|. \end{align*}$$
But $|f(2a + 2b)| = |f((a + b) + (a + b))| = |2f(a + b)| = 2|f(a + b)| \neq 0$. Therefore, $2|f(a)| = |f(2a + 2b)| = 2|f(a + b)|$ and thus $|f(a + b)| = |f(a)|$. Similarly, we can show that $|f(a + b)| = |f(b)|$. So we have $|f(a)| = |f(b)|$. That is $f(a) = f(b)$ or $f(a) = -f(b)$. If $f(a) = f(b)$, then $2|f(a)| = |f(a) + f(a)| = |f(a) + f(b)| = |f(a+b)| = |f(a)|$ and thus $f(a) = 0$ which implies that $f(a + b) = -f(a) - f(b) = -2f(a) = 0$, a contradiction. On the other hand, if $f(a) = -f(b)$, then $0 = |f(a) + f(b)| = |f(a+b)|$, again a contradiction. So in any case we get a contradiction. $\square$