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\pi/ 3 <= (\alpha a +\beta b + \gamma c)/(a+b+c) < \pi}/2

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

May 25, 2021
geometrygeometric inequalityUkrainian TYMalgebraInequality

Problem Statement

Let a,ba, b, and cc be the lengths of the sides of an arbitrary triangle, and let α,β\alpha,\beta, and γ\gamma be the radian measures of its corresponding angles. Prove that π3≤αa+βb+γca+b+c<π2. \frac{\pi}{3}\le \frac{\alpha a +\beta b + \gamma c}{a+b+c} < \frac{\pi}{2}. Suggest spatial analogues of this inequality.
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