埴充4  
索迪公式
http://lyingheart6174.pixnet.net ... %28soddy-formula%29

\(
\begin{array}{l}
(1)因為 a_1  = \sqrt 3 , \\
\displaystyle \sqrt 3  \le a_2  = \sqrt {3 + \sqrt 3 }  \le \frac{{1 + \sqrt {13} }}{2} \\
\displaystyle  \sqrt 3  \le a_3  = \sqrt {3 + \sqrt {3 + \sqrt 3 } }  \le \sqrt {3 + \frac{{1 + \sqrt {13} }}{2}}  = \frac{{1 + \sqrt {13} }}{2} \\
{\rm{          }}..... \\
\displaystyle  \sqrt 3  \le a_n  = \sqrt {3 + \sqrt {3 + ...\sqrt {3 + \sqrt 3 } } }  \le \sqrt {3 + \frac{{1 + \sqrt {13} }}{2}} = \frac{{1 + \sqrt {13} }}{2} \\
\displaystyle  \forall n \in N ,\sqrt 3 \le a_n \le \frac{{1 + \sqrt {13} }}{2}  \\
\end{array}
\)
\(
\begin{array}{l}
(2) \\
\displaystyle  a_{n + 1}^2 - a_n^2 = 3 + a_n - a_n^2 = - (a_n - \frac{{1 + \sqrt {13} }}{2})(a_n - \frac{{1 - \sqrt {13} }}{2}) > 0 \\
因此為遞增數列
\end{array}
\)
(3)
\(
\begin{array}{l}
根據實數完備性，極限值存在，可令 \displaystyle \mathop {\lim }\limits_{n \to \infty } a_n = \alpha  \\
\displaystyle \alpha = \sqrt {3 + \alpha } \to \alpha ^2 = 3 + \alpha \to \alpha = \frac{{1 + \sqrt {13} }}{2} (負不合) \\
\end{array}
\)

[ 本帖最後由 eyeready 於 2017-5-27 19:06 編輯 ]

7(3)
\(
\begin{array}{l}
\displaystyle \sin \left( {\frac{{\frac{{A + B + C}}{2}}}{3}} \right) \ge \frac{{\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}}}{3} \ge \sqrt[3]{{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}} \\
\displaystyle \sin 30^ \circ   \ge \sqrt[3]{{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}} \to \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \le \frac{1}{8} \\
\end{array}
\)