物理

[TINY NSD]三角系列续集的续集

${2.15~\sin^2A+\sin^2B+\sin^2C \le\frac{9}{4}}$

${[证法1]由1.7及2.12，\sin^2A+\sin^2B+\sin^2C=2+2\cos A \cos B \cos C \le 2+2 \cdot \frac{1}{8}=\frac{9}{4}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${[证法2]（来自数之谜）设z_1=1,z_2=e^{2iA}, z_3=e^{2i(A+B)}=3^{-2iC}}$

${由几何关系，易知|z_1|=|z_2|=|z_3|=1，且|z_1-z_2|=2\sin A, |z_2-z_3|=2\sin B, |z_3-z_1|=2\sin C}$

${从而\sin^2 A+\sin^2B+\sin^2 C =\frac{|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2}{4}}$

${而|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2=(z_1-z_2)(\overline{z_1}-\overline{z_2})+(z_2-z_3)(\overline{z_2}-\overline{z_3})+(z_3-z_1)(\overline{z_3}-\overline{z_1})}$

${=2(|z_1\overline{z_1}+|z_2|\overline{z_2}+|z_3|\overline{z_3})-(z_1\overline{z_2}+z_2\overline{z_1}+z_2\overline{z_3}+z_3\overline{z_2}+z_3\overline{z_1}+z_1\overline{z_3})}$

${=3(z_1\overline{z_1}+z_2\overline{z_2}+z_3\overline{z_3})-(z_1+z_2+z_3)(\overline{z_1}+\overline{z_2}+\overline{z_3})}$

${=3(|z_1|^2+|z_2|^2+|z_3|^2)-|z_1+z_2+z_3|^2=9-|z_1+z_2+z_3|^2\le 9}$

${故\sin^2 A+\sin^2B+\sin^2 C \le \frac{9}{4}，取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.16~\cos^2A+\cos^2B+\cos^2C \ge \frac{3}{4}}$

${[证明]由1.6及2.12，\cos^2A+\cos^2B+\cos^2C=1-2\cos A \cos B \cos C \ge 1-2 \cdot \frac{1}{8}=\frac{3}{4}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$