Prove the identity of a triangle in the given problem

[chwala]

Homework Statement

Prove that in a triangle ABC

$\dfrac{1}{a} \cos^2 \dfrac{1}{2} A+ \dfrac{1}{b} \cos^2 \dfrac{1}{2} B +\dfrac{1}{c} \cos^2 \dfrac{1}{2} C = \dfrac{(a+b+c)^2}{abc}$

Relevant Equations

Trig. identities

My approach,
$\cos^2 \frac{1}{2} A = \dfrac{s(s-a)}{bc}$ ...
then it follows that,

$\dfrac{1}{a} \cos^2 \dfrac{1}{2} A+ \dfrac{1}{b} \cos^2 \dfrac{1}{2} B +\dfrac{1}{c} \cos^2 \dfrac{1}{2} C =\dfrac{s(s-a)+s(s-b)+s(s-c)}{abc}$

$=\dfrac {3s^2-s(a+b+c)}{abc}$

$=\dfrac{3s^2-2s^2}{abc}$

$=\dfrac{s^2}{abc}=\dfrac{(a+b+c)^2}{4abc}$

any insight or better approach welcome.

 

[anuttarasammyak]

\frac{1}{a}\cos ^2\frac{A}{2}=\frac{\cos A}{2a}+\frac{1}{2a}=\frac{b^2+c^2-a^2}{4abc}+\frac{1}{2a}
using cosine theorem
a^2=b^2+c^2-2bc \cos A
Doing the same for b and c, the sum is
\frac{a^2+b^2+c^2}{4abc}+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=\frac{(a+b+c)^2}{4abc}

 

[Gavran]

$\begin{align} \cos^2\frac A2&=\frac {s(s-a)}{bc}\nonumber\\ &=\frac{(a+b+c)(b+c-a)}{4bc}\nonumber\\ &=\frac{ab+b^2+bc+ac+bc+c^2-a^2-ab-ac}{4bc}\nonumber\\ &=\frac{b^2+c^2-a^2}{4bc}+\frac 12\nonumber \end{align}$

The proof in the post #1 and the proof in the post #2 are almost the same. They use the same principle.