Proving a trigonometric inequality in a triangle In $\Delta ABC$, prove that $$\sin\left(\frac{\pi-A}{4}\right) \sin\left(\frac{\pi-B}{4}\right)\sin\left(\frac{\pi-C}{4}\right) \geq \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ Some ideas on how to start manipulating. I tried breaking it in sum, but it led to nothing. Thanks.

Hint: In $\triangle{ABC}$ we have,
\begin{align*}\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}} &=1+4\sin{\left(\frac{B+C}{4}\right)}\sin{\left(\frac{A+C}{4}\right)}\sin{\left(\frac{A+B}{4}\right)}\\
\cos{A}+\cos{B}+\cos{C}&=1+4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}.\end{align*}
