67th Romanian Mathematical Olympiad

Plugging $a=b=0$ yields $f^2(0)\le 0$, hence $f(0)=0$. Again, plugging $b=0$, and then $a=0$, we obtain $f(a^2)\le af(a)$, $\forall a\in \mathbb{R}$ and $f(b^2)\ge bf(b)$, $\forall b\in \mathbb{R}$, therefore $f(x^2)=xf(x)$, $\forall x\in \mathbb{R}$. Replacing the latter in the given equation, we get $f(a)f(b)\le ab$, $\forall a,b\in \mathbb{R}$. Also, $-xf(-x)=f((-x{)}^2)=f(x^2)=xf(x)$, hence $f$ is an odd function. The last two relations give $f(a)f(b)=-f(a)f(-b)\ge -(-ab)=ab$, so $f(a)f(b)=ab$, $\forall a,b\in \mathbb{R}$. Thus, $f^2(1)=1$, which implies $f(1)=\pm 1$ and either $f(x)=x,\forall x\in \mathbb{R}$, or $f(x)=-x,\forall x\in \mathbb{R}$.

It is easy to check that both functions are solutions of the problem.