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 $\alpha=\beta=\gamma\ne\frac{k\pi}6$