回復 9# panda.xiong 的帖子
若偷用第一題的結論以及 \(rs=\sqrt{s(s-a)(s-b)(s-c)}\)
\(\displaystyle \frac{1}{(s-a)^2}+\frac{1}{(s-b)^2}+\frac{1}{(s-c)^2}\geq\frac{s}{(s-a)(s-b)(s-c)}=\frac{1}{r^2}\)
若沒有第一題的話,原命題
\(\displaystyle \Leftrightarrow \Big(\frac{r}{s-a}\Big)^2+\Big(\frac{r}{s-b}\Big)^2+\Big(\frac{r}{s-c}\Big)^2\geq 1\)
\(\displaystyle \Leftrightarrow \tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\geq 1\)
由以下柯西不等式以及 \(\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\geq 0\),可知上式成立
\(\displaystyle \Big(\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\Big)\Big(\tan^2\frac{B}{2}+\tan^2\frac{C}{2}+\tan^2\frac{A}{2}\Big)\geq\Big(\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}\Big)^2=1\)
最後的等號利用了 \(\displaystyle \frac{A+B+C}{2}=\frac{\pi}{2} \Rightarrow \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1\) 這件事