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c.Ta có:$\cos\dfrac{A}{2}+\cos\dfrac{B}{2}=2\cos(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\cos(\dfrac{A}{4}+\dfrac{B}{4})$$\cos\dfrac{C}{2}+\cos\dfrac{\pi}{6}=2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})$$\cos(\dfrac{A}{4}+\dfrac{B}{4})+\cos(\dfrac{C}{4}+\dfrac{\pi}{12})=2\cos\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\cos\dfrac{\pi}{6}$Suy ra: $\cos\dfrac{A}{2}+\cos\dfrac{B}{2}+\cos\dfrac{C}{2}\le3\cos\dfrac{\pi}{6}=\dfrac{3\sqrt3}{2}$Ta có: $\dfrac{\cos\dfrac{A}{2}}{1+\cos A}+\dfrac{\cos\dfrac{B}{2}}{1+\cos B}+\dfrac{\cos\dfrac{C}{2}}{1+\cos C}$$=\dfrac{1}{2\cos\dfrac{A}{2}}+\dfrac{1}{2\cos\dfrac{B}{2}}+\dfrac{1}{2\cos\dfrac{C}{2}}$$\ge\dfrac{9}{2\cos\dfrac{A}{2}+2\cos\dfrac{B}{2}+2\cos\dfrac{C}{2}}\ge\sqrt3$Dấu bằng xảy ra khi: $\Delta ABC$ đều.

c.Ta có:$\cos\frac{A}{2}+\cos\frac{B}{2}=2\cos(\frac{A}{4}+\frac{B}{4})\cos(\frac{A}{4}-\frac{B}{4})\le2\cos(\frac{A}{4}+\frac{B}{4})$$\cos\frac{C}{2}+\cos\frac{\pi}{6}=2\cos(\frac{C}{4}+\frac{\pi}{12})\cos(\frac{C}{4}-\frac{\pi}{12})\le2\cos(\frac{C}{4}+\frac{\pi}{12})$ $\cos(\frac{A}{4}+\frac{B}{4})+\cos(\frac{C}{4}+\frac{\pi}{12})=2\cos\frac{\pi}{6}\cos(\frac{A}{8}+\frac{B}{8}-\frac{C}{8}-\frac{\pi}{24})\le2\cos\frac{\pi}{6}$Suy ra: $\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\le3\cos\frac{\pi}{6}=\frac{3\sqrt3}{2}$Ta có: $\frac{\displaystyle\cos\frac{A}{2}}{1+\cos A}+\frac{\displaystyle\cos\frac{B}{2}}{1+\cos B}+\frac{\displaystyle\cos\frac{C}{2}}{1+\cos C}$$=\frac{1}{\displaystyle 2\cos\frac{A}{2}}+\frac{1}{\displaystyle 2\cos\frac{B}{2}}+\frac{1}{\displaystyle 2\cos\frac{C}{2}}$$\ge\frac{9}{\displaystyle 2\cos\frac{A}{2}+2\cos\frac{B}{2}+2\cos\frac{C}{2}}\ge\sqrt3$Dấu bằng xảy ra khi: $\Delta ABC$ đều.