引用:
已知 \(0<a,b,c,d<\pi\)
證明
1. \[\sin \left( {\frac{{a + b}}{2}} \right) \ge \frac{{\left( {\sin a + \sin b} \right)}}{2}\]
2. \[\sin \left( {\frac{{a + b + c + d}}{4}} \right) \ge \frac{{\left( {\sin a + \sin b + \sin c + \sin d} \right)}}{4}\]
3. \[\sin \left( {\frac{{a + b + c}}{3}} \right) \ge \frac{{\left( {\sin a + \sin b + \sin c} \right)}}{3}\]
1.
\[\sin a + \sin b = 2\sin \left( {\frac{{a + b}}{2}} \right)\cos \left( {\frac{{a - b}}{2}} \right) \le 2\sin \left( {\frac{{a + b}}{2}} \right) \cdot 1\]
\[ \Rightarrow \frac{{\left( {\sin a + \sin b} \right)}}{2} \le \sin \left( {\frac{{a + b}}{2}} \right)\]
以下，模仿算幾不等式的証明法，
2.
\[\sin \left( {\frac{{a + b + c + d}}{4}} \right) = \sin \left( {\frac{{\frac{{a + b}}{2} + \frac{{c + d}}{2}}}{2}} \right) \ge \frac{{\sin \left( {\frac{{a + b}}{2}} \right) + \sin \left( {\frac{{c + d}}{2}} \right)}}{2}\]
\[ \ge \frac{{\frac{{\sin a + \sin b}}{2} + \frac{{\sin c + \sin d}}{2}}}{2} = \frac{{\sin a + \sin b + \sin c + \sin d}}{4}\]
3.
令 \(d = \frac{{a + b + c}}{3}\) → \(3d=a+b+c\) → \(a+b+c+d = 4d\) → \(\frac{a+b+c+d}{4} = d\)
由 \(\sin \left( {\frac{{a + b + c + d}}{4}} \right) \ge \frac{{\sin a + \sin b + \sin c + \sin d}}{4}\)
可得
\[\sin \left( d \right) \ge \frac{{\sin a + \sin b + \sin c + \sin d}}{4}\]
\[ \Rightarrow 4\sin \left( d \right) \ge \sin a + \sin b + \sin c + \sin d\]
\[ \Rightarrow 3\sin \left( d \right) \ge \sin a + \sin b + \sin c\]
\[ \Rightarrow \sin \left( d \right) \ge \frac{{\sin a + \sin b + \sin c}}{3}\]
\[ \Rightarrow \sin \left( {\frac{{a + b + c}}{3}} \right) \ge \frac{{\sin a + \sin b + \sin c}}{3}.\]
