a+b+c=180∘ prove that cos⁡a+cos⁡b+cos⁡c

If ${\displaystyle a=b=c=60º,\cos a+\cos b+\cos c=3\cdot \frac{1}{2}=1.5.}$
${\displaystyle \cos c=\cos (180-(a+b))=-\cos (a+b).}$
Let

${\displaystyle a=60-a^{\prime }\text{and}b}$

$=60+a^{\prime }f\text{or}0\le a^{\prime }<60º,\cos \left(0\right)=1,\cos \left(60\right)=0.5.$
Trig identity:

${\displaystyle \cos (A-B)+\cos (A+B)}$

$=\cos A\cos +\sin A\sin B+\cos A\cos B-\sin A\sin B=2\cos A\cos B.$
We have:

${\displaystyle \cos (60-a^{\prime })+\cos (60+a^{\prime })-\cos \left(120\right)}$

$=2\cos \left(60\right)\cos \left(a^{\prime }\right)+0.5=\cos \left(a^{\prime }\right)+0.5.\le 1.5.$
Also, as $a+b$ approaches 0, the expression on the left $<1+1-1<1$. 1 is less than 1.5.
As $a+b$ approaches 180, the expression becomes ${\displaystyle \cos a+\cos b+1.}$
Let $a=90-a^{\prime }$ and $b=90+a^{\prime },$

$0{\displaystyle \cos (90-a^{\prime })+\cos (90+a^{\prime })+1=2\cos 90\cos \left(a^{\prime }\right)+1=1}$

less than 1.5.