Question

In any triangle ABC prove that identities.
In a triangle prove that $\cos A+\cos B+\cos C>1$ but not greater than $3/2$.

Answer 1

$\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}>1$ ..$\left(1\right)$
as neither of $\sin \frac{A}{2},\sin \frac{B}{2},\sin \frac{C}{2}$ is $-$ive or zero.
Again $\cos A+\cos B+\cos C$
$=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+1-2{\sin }^2\frac{C}{2}\le 2\sin \frac{C}{2}\cdot 1+1-2{\sin }^2\frac{C}{2}$
$\because 0\le cos\frac{A-B}{2}\le 1$
$=-2[s^2-s-\frac{1}{2}]$, where $s=\sin \frac{C}{2}$
$=-2[{(s-\frac{1}{2})}^2-\frac{1}{2}-\frac{1}{4}]$
$=\frac{3}{2}-{(s-\frac{1}{2})}^2\le \frac{3}{2}$
$\therefore \cos A+\cos B+\cos C\le 3/2$ .$\left(2\right)$
In other words, $\sum \cos A\ngtr 3/2$