$(\cos A+\cos B+\cos C)2abc$

$=\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}\right)2abc$

$=\frac{\left[a(b^2+c^2-a^2)+b\left(a^2+c^2-b^2\right)+c\left(a^2+b^2-c^2\right)\right]2abc}{2abc}$

$=a(b^2+c^2-a^2)+b\left(a^2+c^2-b^2\right)+c\left(a^2+b^2-c^2\right)$

Put $a=b-1 \& c=b+1$ as $a,b,c$ are three consecutive natural numbers & $a<b<c$.

$\therefore (\cos A+\cos B+\cos C)2abc=\left(b-1\right)\left(b^2+{\left(b+1\right)}^2-{\left(b-1\right)}^2\right)+b\left({\left(b-1\right)}^2+{\left(b+1\right)}^2-b^2\right)+\left(b+1\right)\left({\left(b-1\right)}^2+b^2-{\left(b+1\right)}^2\right)$

$=\left(b-1\right)\left(b^2+4b\right)+b\left(2\left(b^2+1\right)-b^2\right)+\left(b+1\right)\left(b^2-4b\right)$

$=3b\left(b^2-2\right)$