[ Extreme points of a triangle ]

Points $A_{1}, B_{1}$ and $C_{1}$ are taken on the sides $B C, C A$ and $A B$ of triangle $A B C$, such that segments $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at a single point $M$. For what position of point $M$ is the value of $\frac{M A_{1}}{A A_{1}} \cdot \frac{M B_{1}}{B B_{1}} \cdot \frac{M C_{1}}{C C_{1}}$ maximized?
Let $\alpha = M A_{1} / A A_{1}, \beta = M B_{1} / B B_{1}$ and $\gamma = M C_{1} / C C_{1}$. Since $\alpha + \beta + \gamma = 1$ (see problem 4.48, a)), then $\sqrt[3]{\alpha \beta \gamma} \leq (\alpha + \beta + \gamma) / 3 = 1 / 3$, and equality is achieved when $\alpha = \beta = \gamma = 1 / 3$, i.e., $M$ is the point of intersection of the medians.