IMC 2002 · Problem 4

In the tetrahedron $OABC$, let $\angle BOC = \alpha$, $\angle COA = \beta$ and $\angle AOB = \gamma$. Let $\sigma$ be the angle between the faces $OAB$ and $OAC$, and let $\tau$ be the angle between the faces $OBA$ and $OBC$. Prove that

$$\gamma > \beta \cdot \cos \sigma + \alpha \cdot \cos \tau.$$