In a triangle ABC, $\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}$

$\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}=1+4\sin \frac{\mathrm{A}}{2}\sin \frac{\mathrm{B}}{2}\sin \frac{\mathrm{C}}{2}>1$
as neither of $\sin \frac{\mathrm{A}}{2},\sin \frac{\mathrm{B}}{2},\sin \frac{\mathrm{C}}{2}$ is – ve or zero
again $\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}$
$1+4\sin \frac{\mathrm{A}}{2}\sin \frac{\mathrm{B}}{2}\sin \frac{\mathrm{C}}{2}\le \frac{3}{2}\Rightarrow \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}\le \frac{3}{2}$