Prove that $\displaystyle 1 <\cos A+\cos B+\cos C$ Prove that $\displaystyle 1 < \cos A+\cos B+\cos C \leq \frac{3}{2}$ where $A+B+C = \pi$
Attempt: $\cos A+\cos B+\cos C= 1+4\cos A\cdot \cos B\cdot \cos C$
Now $\displaystyle \cos A\cdot \cos B\cdot \cos C=\frac{1}{2}\cos A\left[\cos (B-C)-\cos A\right]\leq \frac{1}{2}\cos A(1-\cos A)\leq \frac{1}{2}\left[\frac{\cos A+1-\cos A}{2}\right]^2 = \frac{1}{8}$
so $\displaystyle \cos A+\cos B+\cos C\leq 1+\frac{1}{2} = \frac{3}{2}$
could some help me how to solve for minimum , thanks 

use that $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=1+\frac{r}{R}$$
