Đặt \alpha+\beta+\gamma=\delta \Rightarrow \sum \cos(\alpha+\beta)=\sum\cos(\delta-\gamma)=\cos\delta.\sum\cos\alpha+\sin\delta.\sum\sin\alpha
=\sin^2\delta\frac{\sin \alpha+\sin \beta+\sin \gamma }{\sin(\alpha+\beta+\gamma)}+\cos^2\delta\frac{\cos\alpha+\cos\beta+\cos\gamma}{\cos(\alpha+\beta+\gamma)}=m
Ta lại có P=\sum\cos^2(\alpha+\beta)\ge\frac{(\sum\cos(\alpha+\beta))^2}3=\frac{m^2}3
Dấu bằng khi \alpha=\beta=\gamma\ne\frac{k\pi}6