Let the side length of $\bigtriangleup c$ be $\sqrt c$, $\bigtriangleup b$ be $\sqrt b$, and $\bigtriangleup a$ be $\sqrt a$. Then, the area of $\bigtriangleup (\alpha +b+c)=(\sqrt{b}+\sqrt{c})^2=b+c+2\sqrt{bc}$. Then $\bigtriangleup \alpha=2\sqrt{bc}$, and similarly, $\bigtriangleup \beta=2\sqrt{ac}$ and $\bigtriangleup \gamma=2\sqrt{ab}$. So, $\frac{a}{\alpha}=\frac{a^2\sqrt{bc}}{abc}$, and the LHS of the inequality is $\frac{\sum\limits_{cyc} a^2\sqrt{bc}}{abc}$. Multiplying by $\frac{2}3abc$ on both sides, we get $\frac{\sum\limits_{cyc} a^2\sqrt{bc}}3\ge abc$ Using AM-GM with $a^2\sqrt{bc}, b^2\sqrt{ac},$ and $c^2\sqrt{ab}$, we get the inequality desired.