<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] From the cosine rule:
$$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$
So
$$\begin{align}a\cos(A)+b\cos(B)+c\cos(C)&=a\frac{b^2+c^2-a^2}{2bc}+b\frac{a^2+c^2-b^2}{2ac}+c\frac{a^2+b^2-c^2}{2ab}\\
&=a^2\frac{b^2+c^2-a^2}{2abc}+b^2\frac{a^2+c^2-b^2}{2abc}+c^2\frac{a^2+b^2-c^2}{2abc}\\
&=\frac{a^2b^2+a^2c^2-a^4+a^2b^2+b^2c^2-b^4+a^2c^2+b^2c^2-c^4}{2abc}\\
&=\frac{-(a-b-c)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\
&=\frac{(b+c-a)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\
&=\frac{(2s-2a)(2s-2c)(2s-2b)(2s)}{2abc}\\
&=\frac{8(s-a)(s-c)(s-b)(s)}{abc}\\
&=\frac{8\Delta^2}{abc}\\
\end{align}$$
</s>