From $\cos A=\frac{b^2+c^2-a^2}{2bc}$ and the identity $\cos A=1-2\sin^2 \frac{A}{2}$, we have $2\sin^2 \frac{A}{2}=\frac{a^2-(b-c)^2}{2bc}$. Thus, $4\sin^2 \frac{A}{2}=\frac{a^2-(b-c)^2}{bc}\leq \frac{a^2}{bc}$ (since $(b-c)^2\geq 0)$. Similarly, $4\sin^2 \frac{B}{2}\leq \frac{b^2}{ca}$ and $4\sin^2 \frac{C}{2}\leq \frac{c^2}{ab}$. Therefore, adding these three inequalities, we get the desired inequality.

We know that$:$ $$\sin(\frac{\alpha}{2})=\sqrt{\frac{1-cos(\alpha)}{2}}$$Also$:cosA=\frac{b^2{2}+c^{2}-a^{2}}{2bc};cosB=\frac{c^{2}+a^{2}-b^{2}}{2ca};cosC=\frac{a^{2}+b^{2}-c^{2}}{2ab}$ So we have to prove that$: \sum \frac{a^{2}}{bc}\geq 4(\frac{3}{2}-\sum \frac{a^{2}+b^{2}-c^{2}}{4ab})\Leftrightarrow \sum \frac{b^{2}+c^{2}}{2bc}\geq 6$ Which is just $AM-GM$ $\blacksquare$