给定正整数 $ n$，求最小的正数 $\lambda$，使得对于任何 $\theta_i\in\left(0, \frac{\pi}{2}\right)(i=1,2, \cdots, n)$，只要 $\tan \theta_{1} \cdot \tan \theta_{2} \cdots \cdots \cdot \tan \theta_{n}=2^{\tfrac{n}{2}}$，就有 $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n}$ 不大于 $\lambda$．
【难度】
【出处】
2003第18届CMO试题
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【答案】
略
【解析】
d当 $n= 1$ 时 $\cos \theta_{1}=\left(1+\tan ^{2} \theta_{1}\right)^{-\tfrac{1}{2}}=\dfrac{\sqrt{3}}{3}$ 有 $\lambda=\dfrac{\sqrt{3}}{3}$．当 $n=2$ 时，可以证明
$\cos \theta_{1}+\cos \theta_{2} \leqslant \dfrac{2 \sqrt{3}}{3}$ ①
且当 $\theta_{1}=\theta_{2}=\arctan \sqrt{2}$ 时等号成立．事实上式 ① $\Leftrightarrow \cos ^{2} \theta_{1}+\cos ^{2} \theta_{2}+2 \cos \theta_{1} \cos \theta_{2} \leqslant \dfrac{4}{3}$ 即
$\dfrac{1}{1+\tan ^{2} \theta_{1}}+\dfrac{1}{1+\tan ^{2} \theta_{2}}+2 \sqrt{\dfrac{1}{\left(1+\tan ^{2} \theta_{1}\right)}\left(1+\tan ^{2} \theta_{2}\right)} \leqslant \dfrac{4}{3}$ ②
由 $\tan \theta_{1} \tan \theta_{2}=2$ 可得式 ② $\Leftrightarrow \dfrac{2+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}+2 \sqrt{\dfrac{1}{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}} \leqslant \dfrac{4}{3}$ ③
记 $x=\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}$，则式 ③ $\Leftrightarrow 2 \sqrt{\dfrac{1}{5+x}} \leqslant \dfrac{14+x}{3(5+x)}$
即 $36(5+x) \leqslant 196+28 x+x^{2}$ ④
显然式 ④ $\Leftrightarrow x^{2}-8 x+16=(x-4)^{2} \geqslant 0$，于是，$\lambda=\dfrac{2 \sqrt{3}}{3}$．
当 $n\geqslant 3$ 时，不妨设 $\theta_{1} \geqslant \theta_{2} \geqslant \cdots \geqslant \theta_{n}$，则 $\tan \theta_{1} \tan \theta_{2} \tan \theta_{3} \geqslant 2 \sqrt{2}$ 由于 $\cos \theta_{i}=\sqrt{1-\sin ^{2} \theta_{i}}<1-\dfrac{1}{2} \sin ^{2} \theta_{i}$ 则 $\cos \theta_{2}+\cos \theta_{3}<2-\dfrac{1}{2}\left(\sin ^{2} \theta_{2}+\sin ^{2} \theta_{3}\right)<2-\sin \theta_{2} \sin \theta_{3}$ 由 $\tan ^{2} \theta_{1} \geqslant \dfrac{8}{\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}$ 有 $\dfrac{1}{\cos ^{2} \theta_{1}} \geqslant \dfrac{8+\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}{\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}$ 即 $\begin{aligned} \cos \theta_{1} \leqslant & \dfrac{\tan \theta_{2} \tan \theta_{3}}{\sqrt{8+\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}}= \dfrac{\sin \theta_{2} \sin \theta_{3}}{\sqrt{8 \cos ^{2} \theta_{2} \cos ^{2} \theta_{3}+\sin ^{2} \theta_{2} \sin ^{2} \theta_{3}}} \end{aligned}$
于是 $\cos \theta_{2}+\cos \theta_{3}+\cos \theta_{1}<2-\sin \theta_{2} \sin \theta_{3}\left(1-\dfrac{1}{\sqrt{8 \cos ^{2} \theta_{2} \cos ^{2} \theta_{3}+\sin ^{2} \theta_{2} \sin ^{2} \theta_{3}}}\right)$ ⑤
注意到
$8 \cos ^{2} \theta_{2} \cos ^{2} \theta_{3}+\sin ^{2} \theta_{2} \sin ^{2} \theta_{3} \geqslant 1 \Leftrightarrow 8+\tan ^{2} \theta_{2} \tan ^{2} \theta_{3} \geqslant\dfrac{1}{\cos ^{2} \theta_{2} \cos ^{2} \theta_{3}}=\left(1+\tan ^{2} \theta_{2}\right)\left(1+\tan ^{2} \theta_{3}\right) \Leftrightarrow\tan ^{2} \theta_{2}+\tan ^{2} \theta_{3} \leqslant 7 ⑥ $
由此可得当式 ⑥ 成立时，有 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}<2$ ⑦
若式 ⑥ 不成立，则 $\tan ^{2} \theta_{2}+\tan ^{2} \theta_{3}>7$ 因此 $\tan ^{2} \theta_{1} \geqslant \tan ^{2} \theta_{2}>\dfrac{7}{2}$ 所以 $\cos \theta_{1} \leqslant \cos \theta_{2}<\sqrt{\dfrac{1}{1+\dfrac{7}{2}}}=\dfrac{\sqrt{2}}{3}$ 于是 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}<\dfrac{2 \sqrt{2}}{3}+1<2$ 即式 ⑦ 亦成立．
由此可得 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}+\cdots+\cos \theta_{n}<n-1$
另一方面，取 $\theta_{2}=\theta_{3}=\cdots=\theta_{n}=\alpha>0, \alpha \rightarrow 0$，则 $\theta_{1}=\arctan \dfrac{2^\tfrac{n}{2}}{(\tan \alpha)^{n-1}}$
显然 $\theta_{1} \rightarrow \dfrac{\pi}{2}$，从而 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}+\cdots+\cos \theta_{n} \rightarrow n-1$
综上可得 $\lambda＝n- 1$．
$\cos \theta_{1}+\cos \theta_{2} \leqslant \dfrac{2 \sqrt{3}}{3}$ ①
且当 $\theta_{1}=\theta_{2}=\arctan \sqrt{2}$ 时等号成立．事实上式 ① $\Leftrightarrow \cos ^{2} \theta_{1}+\cos ^{2} \theta_{2}+2 \cos \theta_{1} \cos \theta_{2} \leqslant \dfrac{4}{3}$ 即
$\dfrac{1}{1+\tan ^{2} \theta_{1}}+\dfrac{1}{1+\tan ^{2} \theta_{2}}+2 \sqrt{\dfrac{1}{\left(1+\tan ^{2} \theta_{1}\right)}\left(1+\tan ^{2} \theta_{2}\right)} \leqslant \dfrac{4}{3}$ ②
由 $\tan \theta_{1} \tan \theta_{2}=2$ 可得式 ② $\Leftrightarrow \dfrac{2+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}+2 \sqrt{\dfrac{1}{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}} \leqslant \dfrac{4}{3}$ ③
记 $x=\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}$，则式 ③ $\Leftrightarrow 2 \sqrt{\dfrac{1}{5+x}} \leqslant \dfrac{14+x}{3(5+x)}$
即 $36(5+x) \leqslant 196+28 x+x^{2}$ ④
显然式 ④ $\Leftrightarrow x^{2}-8 x+16=(x-4)^{2} \geqslant 0$，于是，$\lambda=\dfrac{2 \sqrt{3}}{3}$．
当 $n\geqslant 3$ 时，不妨设 $\theta_{1} \geqslant \theta_{2} \geqslant \cdots \geqslant \theta_{n}$，则 $\tan \theta_{1} \tan \theta_{2} \tan \theta_{3} \geqslant 2 \sqrt{2}$ 由于 $\cos \theta_{i}=\sqrt{1-\sin ^{2} \theta_{i}}<1-\dfrac{1}{2} \sin ^{2} \theta_{i}$ 则 $\cos \theta_{2}+\cos \theta_{3}<2-\dfrac{1}{2}\left(\sin ^{2} \theta_{2}+\sin ^{2} \theta_{3}\right)<2-\sin \theta_{2} \sin \theta_{3}$ 由 $\tan ^{2} \theta_{1} \geqslant \dfrac{8}{\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}$ 有 $\dfrac{1}{\cos ^{2} \theta_{1}} \geqslant \dfrac{8+\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}{\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}$ 即 $\begin{aligned} \cos \theta_{1} \leqslant & \dfrac{\tan \theta_{2} \tan \theta_{3}}{\sqrt{8+\tan ^{2} \theta_{2} \tan ^{2} \theta_{3}}}= \dfrac{\sin \theta_{2} \sin \theta_{3}}{\sqrt{8 \cos ^{2} \theta_{2} \cos ^{2} \theta_{3}+\sin ^{2} \theta_{2} \sin ^{2} \theta_{3}}} \end{aligned}$
于是 $\cos \theta_{2}+\cos \theta_{3}+\cos \theta_{1}<2-\sin \theta_{2} \sin \theta_{3}\left(1-\dfrac{1}{\sqrt{8 \cos ^{2} \theta_{2} \cos ^{2} \theta_{3}+\sin ^{2} \theta_{2} \sin ^{2} \theta_{3}}}\right)$ ⑤
注意到
$8 \cos ^{2} \theta_{2} \cos ^{2} \theta_{3}+\sin ^{2} \theta_{2} \sin ^{2} \theta_{3} \geqslant 1 \Leftrightarrow 8+\tan ^{2} \theta_{2} \tan ^{2} \theta_{3} \geqslant\dfrac{1}{\cos ^{2} \theta_{2} \cos ^{2} \theta_{3}}=\left(1+\tan ^{2} \theta_{2}\right)\left(1+\tan ^{2} \theta_{3}\right) \Leftrightarrow\tan ^{2} \theta_{2}+\tan ^{2} \theta_{3} \leqslant 7 ⑥ $
由此可得当式 ⑥ 成立时，有 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}<2$ ⑦
若式 ⑥ 不成立，则 $\tan ^{2} \theta_{2}+\tan ^{2} \theta_{3}>7$ 因此 $\tan ^{2} \theta_{1} \geqslant \tan ^{2} \theta_{2}>\dfrac{7}{2}$ 所以 $\cos \theta_{1} \leqslant \cos \theta_{2}<\sqrt{\dfrac{1}{1+\dfrac{7}{2}}}=\dfrac{\sqrt{2}}{3}$ 于是 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}<\dfrac{2 \sqrt{2}}{3}+1<2$ 即式 ⑦ 亦成立．
由此可得 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}+\cdots+\cos \theta_{n}<n-1$
另一方面，取 $\theta_{2}=\theta_{3}=\cdots=\theta_{n}=\alpha>0, \alpha \rightarrow 0$，则 $\theta_{1}=\arctan \dfrac{2^\tfrac{n}{2}}{(\tan \alpha)^{n-1}}$
显然 $\theta_{1} \rightarrow \dfrac{\pi}{2}$，从而 $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}+\cdots+\cos \theta_{n} \rightarrow n-1$
综上可得 $\lambda＝n- 1$．
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