In a triangle ABC, (b + c) cos A + (c + a) cos B + (a + b) cos C is equal to In a triangle $ABC$ 

$$(b + c)\cos A + (c + a)\cos B + (a + b)\cos C=?$$
we know 

$$a = b \cos C + c \cos B$$
$$b= a \cos C + c \cos A$$ 
$$c= b \cos A + a \cos B$$ 
adding together we get 
$$(a+b) \cos C + (b+c) \cos A + (c+a) \cos B =a +b+c $$
If we write: 


$b+c=2S-a$ 
$c+a = 2S-b$
$a+b= 2S -c$

Then by expanding the bracket and taking $2S$ as common factor we can write $2S(\cos a +\cos b +\cos c)-(a\cos a+b\cos b+c\cos c)$
Now since $a+b+c=180 \cos a +\cos b +\cos c = 1+$ 4\sin a/2.sinb/2.sinc/2 and acosa+bcosb +ccosc is $S$.
Therefore solving further we get $2S(1+4 \sin a/2\sin b/2\sin c/2)-S = S + 4 r/R$  where $r$ is inradius and $R$ is circumradius as $4 \sin a/2 \sin b/2\ sin c/2 is r/R$ . 


I'm stuck up after here.   

Help me if I'm doing something wrong