Prove triangle ABC is equilateral triangle given that $2\sin A+3\sin B+4\sin C = 5\cos\frac{A}{2} + 3\cos\frac{B}{2} + \cos\frac{C}{2}$

Question

Let the triangle ABC have angles such that:

$2\sin A+3\sin B+4\sin C = 5\cos\frac{A}{2} + 3\cos\frac{B}{2} + \cos\frac{C}{2}$

Prove triangle ABC is equilateral triangle.

Answer 1

$10\cos\dfrac{A}{2} + 6\cos\dfrac{B}{2} + 2\cos\dfrac{C}{2} = 10\sin \left(\dfrac{B+C}{2} \right) + 6\sin\left(\dfrac{A+C}{2}\right) + 2\sin \left(\dfrac{A+B}{2}\right)$

$\ge 10\sin\left(\dfrac{B+C}{2}\right) \cos\left( \dfrac{B-C}{2}\right) + 6\sin\left(\dfrac{A+C}{2} \right)\cos \left(\dfrac{A-C}{2}\right)+ 2\sin\left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)$

$=5 (\sin B+\sin C)+3 (\sin A+\sin C)+ (\sin A+\sin B) = 4 \sin A + 6 \sin B+ 8 \sin C$

Thus we have $5\cos\dfrac{A}{2} + 3\cos\dfrac{B}{2} + \cos\dfrac{C}{2} \ge 2 \sin A + 3 \sin B+ 4 \sin C$

Its clear that equality occurs when $A=B=C = 60^{\circ}$