Mongolian Mathematical Olympiad

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

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 $\forall x\in \mathbb{R}:g(3x)\le g(x)\Rightarrow g(x_n)\le g(x_{n+1}),\forall n\in \mathbb{N}$. From this follows $g(x_n)\ge g(x_0)=M,\forall n\in \mathbb{N}$ and $x_n\in [-a,a],\forall n\in \mathbb{N}\Rightarrow g(x_n)\le M$ and $\forall n\in \mathbb{N}:g(x_n)=M$.

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