Problem

Source: VII All-Ukrainian Tournament of Young Mathematicians, Qualifying p12

Tags: geometry, geometric inequality, Ukrainian TYM, algebra, Inequality



Let $a, b$, and $c$ be the lengths of the sides of an arbitrary triangle, and let $\alpha,\beta$, and $\gamma$ be the radian measures of its corresponding angles. Prove that $$ \frac{\pi}{3}\le \frac{\alpha a +\beta b + \gamma c}{a+b+c} < \frac{\pi}{2}.$$Suggest spatial analogues of this inequality.