回復 4# doordie25 的帖子
多背點公式吧
\(\displaystyle \cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}} \)
\(\displaystyle \cos\frac{B}{2}=\sqrt{\frac{s(s-b)}{ac}} \)
\(\displaystyle \cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}} \)
\(\displaystyle \tan\frac{A}{2}=\frac{r}{s-a} \)
\(\displaystyle \tan\frac{B}{2}=\frac{r}{s-b} \)
\(\displaystyle \tan\frac{C}{2}=\frac{r}{s-c} \)
\(\displaystyle \frac{\sqrt{\tan\frac{B}{2} \tan\frac{C}{2}}}{\cos\frac{A}{2}}+\frac{\sqrt{\tan\frac{C}{2} \tan\frac{A}{2}}}{\cos\frac{B}{2}}+\frac{\sqrt{\tan\frac{A}{2} \tan\frac{B}{2}}}{\cos\frac{C}{2}} \)
\(\displaystyle =\frac{r \sqrt{bc}}{\sqrt {s(s-a)(s-b)(s-c)}}+\frac{r \sqrt{ac}}{\sqrt {s(s-a)(s-b)(s-c)}}+\frac{r \sqrt{ab}}{\sqrt {s(s-a)(s-b)(s-c)}} \)
\(\displaystyle =\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{s} \)
又 \(\displaystyle a+b \ge 2\sqrt{ab} \)
\(\displaystyle b+c \ge 2\sqrt{bc} \)
\(\displaystyle a+c \ge 2\sqrt{ac} \)
三式相加得到
\(\displaystyle 2(a+b+c) \ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}) \)
故\(\displaystyle \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{s} \le 2 \)
[ 本帖最後由 老王 於 2012-8-2 11:13 AM 編輯 ]
