<s>[INST] In a triangle, prove that $a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$ I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$
where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS:
$$
\begin {align}
a\cos A&+b\cos B+c\cos C\\ &= a\left[2\cos^2\left(\frac A2\right)-1\right]+b\left[2\cos^2\left(\frac B2\right)-1\right]+c\left[2\cos^2\left(\frac C2\right)-1\right]\\
&=a\left[2 \frac{s(s-a)}{bc} -1\right]+b\left[2 \frac{s(s-b)}{ac} -1\right]+c\left[2 \frac{s(s-c)}{ab} -1\right]\\
&=2a\left (\frac{s(s-a)}{bc}\right)+2b\left (\frac{s(s-b)}{ac}\right)+2c\left( \frac{s(s-c)}{ab}\right)-2s\\
&=\frac{2s}{abc}[a^2(s-a)+b^2(s-b)+c^2(s-c)-abc]
\end{align}$$
where $$s=\frac{a+b+c}{2}$$
I want to convert that last equation into Heron's formula: $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
But I'm stuck there. Any help please?
 [/INST] The polynomial $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}$$ vanishes for $c:=a+b$ and because it is cyclic it vanishes also at $a:=b+c$ and at $b:=a+c$. Notice that the same is true for $$(s-a)(s-b)(s-c).$$
In fact, for the second polynomial it is clear and for the first putting $c:=a+b$ in the first polynomial we get 
$$\frac{2ba^2+2ab^2-2ab(a+b)}{8}\equiv 0.$$
Finally the two polynomials have the same leading coefficient $\frac{1}{8}$.
Therefore $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}=(s-a)(s-b)(s-c).$$
</s>