<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] $t=a^2(s-a)+b^2(s-b)+c^2(s-c)-abc=a^2(s-a)+b^2(s-a)+c^2(s-a)+a(b^2+c^2-bc)-(b^3+c^3)\\b^3+c^3=(b+c)(b^2+c^2-bc)\\t=(s-a)(a^2+b^2+c^2)-(b^2+c^2-bc)(b+c-a),b+c-a=2(s-a)\\t=(s-a)(a^2+b^2+c^2-2(b^2+c^2-bc))=(s-a)(a^2-(b^2+c^2-2bc))=(s-a)(a^2-(b-c)^2)=(s-a)(a+c-b)(a+b-c)=4(s-a)(s-b)(s-c)$
</s>