For all $ x \in \left[0, \dfrac{\pi}{2} \right]$ we have: $ \cos 2x \ge -1$ and $ \sin x \ge 0$. Thus, $ \sin x \cos 2x \ge -\sin x$ with equality iff $ x = 0$ OR $ x = \dfrac{\pi}{2}$. Therefore, $ \sin\alpha\cos 2\alpha+\sin\beta\cos 2\beta+\sin\gamma\cos 2\gamma \ge -(\sin\alpha+\sin\beta+\sin\gamma) = -1$ with equality iff $ \alpha,\beta,\gamma \in \left\{ 0 , \dfrac{\pi}{2} \right\}$. Since $ \sin(0) = 0$ and $ \sin\left(\dfrac{\pi}{2}\right) = 1$, one of $ \alpha,\beta,\gamma$ must be equal to $ 1$ and the other two must be equal to $ 0$. Therefore, $ \sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma = 1$.