$\sqrt{\frac{15}4+\sum\cos(A-B)}\ge\sum\sin A$ in a triangle?

Question

How can I prove that ( $\small{\sum}$ denotes cyclic sum here), for any triangle $ABC$: $$\sqrt{\frac{15}4+\sum\cos(A-B)}\ge\sum\sin A$$

I don't see where to begin even. Any hints would be appreciated :)

Answer 1

Squaring both sides leads to,

$\displaystyle \dfrac{15}{4} + \sum\limits_{cyc}\cos(A-B) \ge (\sum\limits_{cyc} \sin A)^2 = \sum\limits_{cyc} \sin^2 A + 2\sum\limits_{cyc}\sin A\sin B$

$\implies \displaystyle \dfrac{15}{4} + \sum\limits_{cyc}(\cos A \cos B + \sin A \sin B) \ge 3 - \sum\limits_{cyc} \cos^2A + 2 \sum\limits_{cyc}\sin A\sin B \implies \sum\limits_{cyc} \cos^2A + \sum\limits_{cyc}(\cos A \cos B - \sin A \sin B) + \dfrac34 \ge 0 \implies \sum\limits_{cyc} \cos^2A + \sum\limits_{cyc}\cos (A+B) + \dfrac34 = \sum\limits_{cyc}\left( \cos^2A - \cos A +\frac14\right)=\sum\limits_{cyc}\left(\cos A - \frac12\right)^2 \ge 0$

which is true.