Prove that $\displaystyle 1 <\cos A+\cos B+\cos C$ [duplicate]

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

marked as duplicate by user228113, Community♦Nov 24 '17 at 19:11

use that $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=1+\frac{r}{R}$$
• then using $R\geq 2r >0$ – DXT Nov 24 '17 at 18:25
This question is as old as the OP's age ( probably )...Let's do the left inequality. $\cos A +\cos B+\cos C > 1\iff \cos A+\cos B -\left(1-\cos C\right) > 0\iff2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)-2\sin^2\left(\frac{C}{2}\right)>0\iff 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)-2\sin^2\left(\frac{C}{2}\right)>0\iff2\sin\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)-\sin\left(\frac{C}{2}\right)\right)>0\iff2\sin\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)-\cos\left(\frac{A+B}{2}\right)\right)>0\iff 4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)>0$. This last inequality is clearly true since each factor is $> 0$. The right inequality is already done.