<s>[INST] Finding all the triangles $ABC$ satisfying $\sum \frac{a^{2}\cos\frac{B-C}{2}}{\sin\frac{A}{2}}=2(a^2+b^2+c^2)$ $\triangle ABC$ has $BC=a, CA=b, AB=c$ and satisfies
$$\dfrac{a^{2}\cos\dfrac{B-C}{2}}{\sin\dfrac{A}{2}}+\dfrac{b^{2}\cos\dfrac{C-A}{2}}{\sin\dfrac{B}{2}}+\dfrac{c^{2}\cos\dfrac{A-B}{2}}{\sin\dfrac{C}{2}}=2(a^2+b^2+c^2)$$ Find all the triangles $ABC$.
 [/INST] Converting all the angles in sides the given condition will look like 
$$\sum \frac{a^{2}\cos\frac{B-C}{2}}{\sin\frac{A}{2}}\frac{2sin(\frac{B+C}{2})}{2cos(\frac{A}{2})}=2(a^2+b^2+c^2)$$ $$\sum \frac{a^{2}(sin{B}+sinC)}{\sin{A}}=2(a^2+b^2+c^2)$$$$\sum \frac{a^{2}(\frac{b}{R}+\frac{c}{R})}{\frac{a}{R}}=2(a^2+b^2+c^2)$$ $$\sum(ab+ac)=2(a^2+b^2+c^2)$$ $$2(ab+bc+ac)=2(a^2+b^2+c^2)$$
Now considering rearrangement inequlity we have$$a^2+b^2+c^2\geq ab+bc+ac $$
now for the equality to hold the condition should be$$a=b=c$$
and you have the triangle.
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