Belarusian Mathematical Olympiad

Answer: $\sqrt[3]{abc}$.
We can consider $x > 0$. Multiplying three inequalities $x \le ap/q$, $x \le bq/r$, $x \le cr/p$, we obtain $x \le \sqrt[3]{abc}$.
It remains to show that the number $\sqrt[3]{abc}$ satisfies the problem condition. It suffices to verify that the system of the equations (with unknown $p, q, r$)
$$\begin{cases} \sqrt[3]{abc} = a \frac{p}{q}, \\ \sqrt[3]{abc} = b \frac{q}{r}, \\ p + q + r = 1 \end{cases}$$
has a positive solution $(p; q; r)$. From the first two equations we present $p$ and $r$ as the functions of $q$ and then replace $p$ and $q$ by these presentations in the third equation. So we find the positive solution
$$p = \frac{yq}{a}, \quad r = \frac{bq}{y}, \quad q = \frac{1}{\left(\frac{y}{a} + \frac{b}{y} + 1\right)}, \quad \text{where} \quad y = \sqrt[3]{abc},$$
as required.