物理 [TINY NSD]三角系列续集的续集
${2.三角不等式（高考党慎入）}$
${2.1~\sin \frac{A}{2}+\sin\frac{B}{2}+\sin \frac{C}{2} \le \frac{3}{2}}$
${[证明]令f(x)=\sin x(0 \lt x \lt \frac{\pi}{2})，则f''(x)=-\sin x \lt 0}$
${又\frac{A}{2},\frac{B}{2}, \frac{C}{2} \in (0, \frac{\pi}{2})且\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}}$
${由Jensen不等式，\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \le 3\sin\frac{\pi}{6} =\frac{3}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.2~\cos \frac{A}{2}+\cos\frac{B}{2}+\cos \frac{C}{2} \le \frac{3\sqrt{3}}{2}}$
${[证明]令f(x)=\cos x(0 \lt x \lt \frac{\pi}{2})，则f''(x)=-\cos x \lt 0}$
${又\frac{A}{2},\frac{B}{2}, \frac{C}{2} \in (0, \frac{\pi}{2})且\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}}$
${由Jensen不等式，\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2} \le 3\cos\frac{\pi}{6} =\frac{3\sqrt{3}}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.3~\tan \frac{A}{2}+\tan\frac{B}{2}+\tan \frac{C}{2} \ge \sqrt{3}}$
${[证明]令f(x)=\tan x(0 \lt x \lt \frac{\pi}{2})，则f''(x)=\frac{2\sin x}{\cos^3 x} \gt 0}$
${又\frac{A}{2},\frac{B}{2}, \frac{C}{2} \in (0, \frac{\pi}{2})且\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}}$
${由Jensen不等式，\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2} \ge 3\tan\frac{\pi}{6} =\sqrt{3}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.4~\sin \frac{A}{2} \sin \frac{B}{2}\sin \frac{C}{2} \le \frac{1}{8}}$
${[证明]由AM-GM不等式与2.1，\sin\frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le (\frac{\sin \frac{A}{2}+\sin\frac{B}{2}+\sin \frac{C}{2}}{3})^3 \le (\frac{1}{2})^3 =\frac{1}{8}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.5~\cos \frac{A}{2} \cos \frac{B}{2}\cos \frac{C}{2} \le \frac{3\sqrt{3}}{8}}$
${[证明]由AM-GM不等式与2.1，\cos\frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \le (\frac{\cos \frac{A}{2}+\cos\frac{B}{2}+\cos \frac{C}{2}}{3})^3 \le (\frac{\sqrt{3}}{2})^3=\frac{3\sqrt{3}}{8}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.6~\tan\frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} \le \frac{\sqrt{3}}{9}}$
${[证明]由1.8与AM-GM不等式，\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} =\sqrt{\tan^2 \frac{A}{2}\tan^2\frac{B}{2}\tan^2\frac{C}{2}}}$
${\le \sqrt{(\frac{\tan\frac{A}{2} \tan\frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2}\tan\frac{A}{2}}{3})^3}=(\frac{1}{3})^\frac{3}{2}=\frac{\sqrt{3}}{9}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.7~\sin A+\sin B+\sin C\le\frac{3\sqrt{3}}{2}}$
${[证法1]令f(x)=\sin x(0 \lt x \lt \pi),则f''(x)=-\sin x\lt 0}$
${而A, B, C \in(0,\pi)且A+B+C=\pi}$
${由Jensen不等式，\sin A+\sin B+\sin C \le 3\sin \frac{\pi}{3}=\frac{3\sqrt{3}}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${[证法2]由2.5及1.1，\sin A+\sin B+\sin C=4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \le 4 \cdot\frac{3\sqrt{3}}{8}=\frac{3\sqrt{3}}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.8~\cos A+\cos B+\cos C \le\frac{3}{2}}$
${[证法1]不妨设A \ge B \ge C，则B \lt \frac{\pi}{2}, C\le\frac{\pi}{3}, A \ge \frac{\pi}{3}}$
${令f(x)=\cos x+\cos(\theta-x), 其中\theta \in (\frac{\pi}{2}, \pi), x\in(0, \theta)}$
${则f'(x)=\sin(\theta-x)-\sin x=2\cos \frac{\theta}{2} \sin (\frac{\theta}{2}-x)}$
${由于\sin\frac{\theta}{2}\lt 0, \cos \frac{\theta}{2} \lt 0, f(x)在(0,\frac{\theta}{2})时单调递增，在(\frac{\theta}{2},\theta)单调递减}$
${令\theta=A+C, 则\frac{\pi}{2} \lt \theta \lt \pi}$
${(1)若\frac{\pi}{2}\lt\theta \le \frac{2\pi}{3}，则A \ge \frac{\pi}{3}\ge\frac{\theta}{2}}$
${故f(A) \le f(\frac{\pi}{3})，即\cos A+\cos C \le \cos\frac{\pi}{3}+\cos (\frac{2\pi}{3}-B)}$
${令\theta=\frac{2\pi}{3}，则f(B) \le f(\frac{\pi}{3})，即\cos B+\cos(\frac{2\pi}{3}-B) \le 2\cos \frac{\pi}{3}}$
${故\cos A+\cos B+\cos C \le 3\cos \frac{\pi}{3} =\frac{3}{2}}$
${(2)若\frac{2\pi}{3}\lt\theta \lt \pi，则C \le \frac{\pi}{3}\lt\frac{\theta}{2}}$
${故f(C) \lt f(\frac{\pi}{3})，即\cos A+\cos C \lt \cos\frac{\pi}{3}+\cos (\frac{2\pi}{3}-B)}$
${令\theta=\frac{2\pi}{3}，则f(B) \le f(\frac{\pi}{3})，即\cos B+\cos(\frac{2\pi}{3}-B) \le 2\cos \frac{\pi}{3}}$
${故\cos A+\cos B+\cos C \lt 3\cos \frac{\pi}{3} =\frac{3}{2}}$
${综上，总有\cos A+\cos B+\cos C \le \frac{3}{2}，取等当且仅当A=B=C=\frac{\pi}{3}}$
${[证法2]不妨设A \ge B \ge C}$
${则0\lt\frac{A+B}{2} \lt \frac{\pi}{2}, \cos \frac{A+B}{2} \gt 0}$
${\cos A+\cos B=2\cos \frac{A+B}{2}\cos\frac{A-B}{2} \le 2\cos \frac{A+B}{2}}$
${令f(x)=\cos x(0\lt x \lt \frac{\pi}{2})，则f''(x)=-\cos x \lt 0}$
${而\frac{A+B}{2},C\in (0, \frac{\pi}{2})}$
${故由Jensen不等式，2\cos \frac{A+B}{2} +\cos C \le 3\cos \frac{\pi}{3}=\frac{3}{2}}$
${从而\cos A+\cos B+\cos C \le 2\cos \frac{A+B}{2}+\cos C \le \frac{3}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${[证法3]由2.4及1.2，\cos A+\cos B+\cos C=4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}+1 \le 4 \cdot\frac{1}{8}+1=\frac{3}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.9~\tan A+\tan B+\tan C \ge 3\sqrt{3} 或 \lt 0}$
${[证明]若\tan A+\tan B+\tan C \ge 0}$
${则由AM-GM不等式，\tan A+\tan B+\tan C \ge 3\sqrt[3]{\tan A\tan B\tan C}=3\sqrt[3]{\tan A+\tan B+\tan C}（由1.3）}$
${解得\tan A+\tan B+\tan C \ge 3\sqrt{3}}$
${故\tan A+\tan B+\tan C \lt 0 或 \ge 3\sqrt{3}，取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.10~\sin A \sin B \sin C \le\frac{3\sqrt{3}}{8}}$
${[证明]由2.7和AM-GM不等式，\sin A\sin B \sin C \le (\frac{\sin A+\sin B+\sin C}{3})^3 \le \frac{3\sqrt{3}}{8}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.11~\tan A \tan B \tan C \ge 3\sqrt{3} 或\lt 0}$
${[证明]由1.5和2.9，\tan A \tan B \tan C =\tan A+\tan B+\tan C \ge 3\sqrt{3} 或 \lt 0}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.12~-1 \lt \cos A\cos B\cos C \le\frac{1}{8}}$
${[证法1]若A,B,C \in (0,\frac{\pi}{2})，由2.8和AM-GM不等式，\cos A \cos B \cos C \le(\frac{\cos A+\cos B+\cos C}{3})^3 \le \frac{1}{8}}$
${否则，不妨设A \ge B \ge C，由于B \lt \frac{\pi}{2}，必有A \ge\frac{\pi}{2}，从而\cos A \cos B \cos C \le 0 \lt\frac{1}{8}}$
${而由于A,B,C \in(0,\pi),有|\cos A|, |\cos B|, |\cos C| \lt 1}$
${故\cos A \cos B\cos C \ge -|\cos A \cos B \cos C| \gt -1}$
${综上，-1 \lt \cos A\cos B\cos C \le \frac{1}{8}，右侧取等当且仅当A=B=C=\frac{\pi}{3}}$
${[证法2]由2.10及2.11：}$
${(1)若\tan A \tan B\tan C \lt 0，则由于\sin A \sin B \sin C \gt 0，有\cos A \cos B \cos C=\frac{\sin A \sin B \sin C}{\tan A \tan B \tan C} \lt 0 \lt \frac{1}{8}}$
${(2)若\tan A \tan B\tan C \ge 3\sqrt{3}，则由于0 \lt \sin A \sin B \sin C \le\frac{3\sqrt{3}}{8}}$
${有\cos A \cos B\cos C=\frac{\sin A \sin B \sin C}{\tan A \tan B \tan C} \le\frac{\frac{3\sqrt{3}}{8}}{3\sqrt{3}}=\frac{1}{8}}$
${(3)若\tan A \tan B\tan C无意义，则三角形为直角三角形，显然\cos A \cos B \cos C =0 \lt\frac{1}{8}}$
${而由于A,B,C \in(0,\pi),有|\cos A|, |\cos B|, |\cos C| \lt 1}$
${故\cos A \cos B\cos C \ge -|\cos A \cos B \cos C| \gt -1}$
${综上，-1 \lt \cos A\cos B\cos C \le \frac{1}{8}，右侧取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.13~\sin 2A+\sin 2B+\sin 2C \le\frac{3\sqrt{3}}{2}}$
${[证明]由1.4及2.10，\sin2A+\sin 2B+\sin 2C =4\sin A \sin B \sin C \le \frac{3\sqrt{3}}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.14~\cos 2A+\cos 2B+\cos 2C \ge-\frac{3}{2}}$
${[证法1]由1.5及2.12，\cos2A+\cos 2B+\cos 2C=-1-4\cos A \cos B \cos C \ge -1-4\cdot \frac{1}{8}=-\frac{3}{2}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${[证法2]不妨设A \ge B \ge C, 则B \lt \frac{\pi}{2}, 2(A+C)\in (\pi, 2\pi),A \ge \frac{\pi}{3} \ge C}$
${令f(x)=\cos x+\cos (\theta-x),其中\theta \in (\pi, 2\pi), x \in (0,\theta)}$
${则f'(x)=\sin(\theta-x)-\sin x=2\cos \frac{\theta}{2} \sin(\frac{\theta}{2}-x)}$
${而\cos\frac{\theta}{2} \lt 0，故f(x)在(0,\frac{\theta}{2})单调递减，(\frac{\theta}{2}, \theta)单调递增}$
${令\theta=2(A+C)}$
${(1)若\pi \lt\theta \le \frac{4\pi}{3}，则2A \ge \frac{2\pi}{3} \ge\frac{\theta}{2}}$
${故\cos 2A+\cos 2C=f(2A) \ge f(\frac{2\pi}{3}) =\cos \frac{2\pi}{3}+\cos (\theta-\frac{2\pi}{3})=\cos\frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B)}$
${又令\theta=\frac{4\pi}{3}，则f(2B) \le f(\frac{2\pi}{3}),\cos(\frac{4\pi}{3}-2B)+ \cos 2B \ge2\cos \frac{2\pi}{3}}$
${故\cos 2A+\cos2B+\cos 2C \ge \cos \frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B) +\cos 2B \ge 3\cos\frac{2\pi}{3}=-\frac{3}{2}}$
${(2)若\frac{4\pi}{3}\lt \theta \lt \pi，则2C \lt \frac{2\pi}{3} \le\frac{\theta}{2}}$
${故\cos 2A+\cos 2C=f(2C) \gt f(\frac{2\pi}{3}) =\cos \frac{2\pi}{3}+\cos (\theta-\frac{2\pi}{3})=\cos\frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B)}$
${又令\theta=\frac{4\pi}{3}，则f(2B) \le f(\frac{2\pi}{3}),\cos(\frac{4\pi}{3}-2B)+ \cos 2B \ge2\cos \frac{2\pi}{3}}$
${故\cos 2A+\cos2B+\cos 2C \gt \cos \frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B) +\cos 2B \ge 3\cos\frac{2\pi}{3}=-\frac{3}{2}}$
${综上，总有\cos 2A+\cos2B+\cos 2C \ge -\frac{3}{2}，取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.15~\sin^2A+\sin^2B+\sin^2C \le\frac{9}{4}}$
${[证法1]由1.7及2.12，\sin^2A+\sin^2B+\sin^2C=2+2\cos A \cos B \cos C \le 2+2 \cdot \frac{1}{8}=\frac{9}{4}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$
${[证法2]（来自数之谜）设z_1=1,z_2=e^{2iA}, z_3=e^{2i(A+B)}=3^{-2iC}}$
${由几何关系，易知|z_1|=|z_2|=|z_3|=1，且|z_1-z_2|=2\sin A, |z_2-z_3|=2\sin B, |z_3-z_1|=2\sin C}$
${从而\sin^2 A+\sin^2B+\sin^2 C =\frac{|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2}{4}}$
${而|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2=(z_1-z_2)(\overline{z_1}-\overline{z_2})+(z_2-z_3)(\overline{z_2}-\overline{z_3})+(z_3-z_1)(\overline{z_3}-\overline{z_1})}$
${=2(|z_1\overline{z_1}+|z_2|\overline{z_2}+|z_3|\overline{z_3})-(z_1\overline{z_2}+z_2\overline{z_1}+z_2\overline{z_3}+z_3\overline{z_2}+z_3\overline{z_1}+z_1\overline{z_3})}$
${=3(z_1\overline{z_1}+z_2\overline{z_2}+z_3\overline{z_3})-(z_1+z_2+z_3)(\overline{z_1}+\overline{z_2}+\overline{z_3})}$
${=3(|z_1|^2+|z_2|^2+|z_3|^2)-|z_1+z_2+z_3|^2=9-|z_1+z_2+z_3|^2\le 9}$
${故\sin^2 A+\sin^2B+\sin^2 C \le \frac{9}{4}，取等当且仅当A=B=C=\frac{\pi}{3}}$
${2.16~\cos^2A+\cos^2B+\cos^2C \ge \frac{3}{4}}$
${[证明]由1.6及2.12，\cos^2A+\cos^2B+\cos^2C=1-2\cos A \cos B \cos C \ge 1-2 \cdot \frac{1}{8}=\frac{3}{4}}$
${取等当且仅当A=B=C=\frac{\pi}{3}}$