第1題
\(\begin{align}
  & \frac{{{n}^{3}}-1}{{{n}^{3}}+1}=\frac{n-1}{n+1}\times \frac{n\left( n+1 \right)+1}{n\left( n-1 \right)+1} \\
&  \\
& \frac{1}{3}\times \frac{2}{4}\times \frac{3}{5}\times \cdots \times \frac{n-1}{n+1}=\frac{2}{n\left( n+1 \right)} \\
& \frac{2\left( 2+1 \right)+1}{2\left( 2-1 \right)+1}\times \frac{3\left( 3+1 \right)+1}{3\left( 3-1 \right)+1}\times \frac{4\left( 4+1 \right)+1}{4\left( 4-1 \right)+1}\times \cdots \times \frac{n\left( n+1 \right)+1}{n\left( n-1 \right)+1}=\frac{n\left( n+1 \right)+1}{3} \\
&  \\
& \prod\limits_{k=2}^{n}{\frac{{{k}^{3}}-1}{{{k}^{3}}+1}}=\frac{2}{n\left( n+1 \right)}\times \frac{n\left( n+1 \right)+1}{3}=\frac{2}{3}\left[ 1+\frac{1}{n\left( n+1 \right)} \right] \\
& \frac{2}{3}\left[ 1+\frac{1}{n\left( n+1 \right)} \right]\ge \frac{401}{600} \\
& \cdots \cdots  \\
\end{align}\)

[ 本帖最後由 thepiano 於 2021-10-24 11:52 編輯 ]

第 7 題
AP - PF >= AP - PE = AE
等號成立於 AP 和 BD 垂直