Mongolian Mathematical Olympiad

Substitute $g(x) = f(x) - x$, $x \in \mathbb{R}$. Considering the given conditions we get $f(2x) = g(2x) + 2x \geq g(x) + 2x \Leftrightarrow g(2x) \geq g(x)$ and $f(3x) = g(3x) + 3x \leq g(x) + 3x \Leftrightarrow g(3x) \leq g(x)$. In other words, $\forall x \in \mathbb{R}: g(3x) \leq g(x) \leq g(2x)$.

Let us consider $\forall a > 0$ and let $m = \min_{x \in [-a,a]} g(x)$ and $M = \max_{x \in [-a,a]} g(x)$. By Weierstrass theorem there exists $x_0 \in [-a, a]$ such that $g(x_0) = M$.

Now define a sequence as follows: $x_n = 3^{-n} x_0$, $n = 1, 2, \dots$. Consequently we get $\forall x \in \mathbb{R}: g(3x) \leq g(x) \Rightarrow g(x_n) \leq g(x_{n+1}), \forall n \in \mathbb{N}$. From this follows $g(x_n) \geq g(x_0) = M, \forall n \in \mathbb{N}$ and $x_n \in [-a, a], \forall n \in \mathbb{N} \Rightarrow g(x_n) \leq M$ and $\forall n \in \mathbb{N}: g(x_n) = M$.

Since $g(x)$ is a continuous function, $\lim_{n \to \infty} (g(x_n)) = g(\lim_{n \to \infty} x_n) = g(0) = 0 = M \Rightarrow \forall x \in [-a, a] : g(x) \leq 0$. Similarly we get $g(2x) \geq g(x), \forall x \in \mathbb{R} \Rightarrow g(x) \geq 0, x \in [-a, a]$. Therefore $\forall x \in [-a, a] : g(x) \equiv 0$. It implies $\forall x \in \mathbb{R} : g(x) \equiv 0$ and $f(x) = x$.