Prove $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$ $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$
I need to solve this trigonometric identity for a trianlge. 
I'm not allowed to use the formula $a+b+c=s$ where 's' is perimeter.
My Try
Using sine rule I was able to simplify,
\begin{align}\text{L. H. S.}& =\frac{(\sin A+\sin B+ \sin C)^2}{\sin^2A+\sin^2B+\sin^2C} \\
&=\frac{4\cos^2\dfrac C2\Bigl(\dfrac{\cos(A-B)}{2}+\sin\frac C2\Bigr)^2}{\sin^2A+\sin^2B+\sin^2C}
\end{align}
I'm unable to simplify further, Please give me hint. Thanks in advance.

Start from the RHS and use the following formulas:
Step 1:
$\cot \frac{A}{2} = \frac{\Delta}{(s-b)(s-c)}$ etc. in the numerator
$\cot A = \frac{R}{abc} (b^2 + c^2 - a^2)$ etc. in the denominator
Step 2:
$R = \frac{abc}{4 \Delta}$
Step 3:
$\Delta^2 = s(s-a)(s-b)(s-c)$
You are done!
