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c.Ta có:$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}=2\sin(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\sin(\dfrac{A}{4}+\dfrac{B}{4})$$\sin\dfrac{C}{2}+\sin\dfrac{\pi}{6}=2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})$$\sin(\dfrac{A}{4}+\dfrac{B}{4})+\sin(\dfrac{C}{4}+\dfrac{\pi}{12})=2\sin\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\sin\dfrac{\pi}{6}$Suy ra: $\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le3\sin\dfrac{\pi}{6}=\dfrac{3}{2}$Suy ra: $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\le\left(\dfrac{\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}}{3}\right)^3\le\dfrac{1}{8}$Ta có: $\left(1+\dfrac{1}{\sin\dfrac{A}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{B}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{C}{2}}\right)$$=1+\left(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\right)+\left(\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{B}{2}\sin\dfrac{C}{2}}+\dfrac{1}{\sin\dfrac{C}{2}\sin\dfrac{A}{2}}\right)+\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}$$\ge1+\dfrac{3}{\sqrt[3]{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}}+\dfrac{3}{\sqrt[3]{\sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}\sin^2\dfrac{C}{2}}}+8\ge27$Dấu bằng xảy ra khi: $\Delta ABC$ đều.

e.Ta có:$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}=2\sin(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\sin(\dfrac{A}{4}+\dfrac{B}{4})$$\sin\dfrac{C}{2}+\sin\dfrac{\pi}{6}=2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})$$\sin(\dfrac{A}{4}+\dfrac{B}{4})+\sin(\dfrac{C}{4}+\dfrac{\pi}{12})=2\sin\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\sin\dfrac{\pi}{6}$Suy ra: $\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le3\sin\dfrac{\pi}{6}=\dfrac{3}{2}$Suy ra: $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\le\left(\dfrac{\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}}{3}\right)^3\le\dfrac{1}{8}$Ta có: $\left(1+\dfrac{1}{\sin\dfrac{A}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{B}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{C}{2}}\right)$$=1+\left(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\right)+\left(\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{B}{2}\sin\dfrac{C}{2}}+\dfrac{1}{\sin\dfrac{C}{2}\sin\dfrac{A}{2}}\right)+\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}$$\ge1+\dfrac{3}{\sqrt[3]{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}}+\dfrac{3}{\sqrt[3]{\sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}\sin^2\dfrac{C}{2}}}+8\ge27$Dấu bằng xảy ra khi: $\Delta ABC$ đều.