China Mathematical Olympiad

When $n=1$, $\cos {\theta }_1=(1+{\tan }^2{\theta }_1{)}^{-\frac{1}{2}}=\frac{\sqrt{3}}{3}$. Hence $\lambda =\frac{\sqrt{3}}{3}$.

When $n=2$, we can prove $$\cos {\theta }_1+\cos {\theta }_2\le \frac{2\sqrt{3}}{3},\stackrel{◯}{1}$$ and when ${\theta }_1={\theta }_2=\arctan \sqrt{2}$, the equality holds. In fact, $$\stackrel{◯}{1}\iff {\cos }^2{\theta }_1+{\cos }^2{\theta }_2+2\cos {\theta }_1\cdot \cos {\theta }_2\le \frac{3}{4},$$ $$\text{that is, }\frac{1}{1+{\tan }^2{\theta }_1}+\frac{1}{1+{\tan }^2{\theta }_2}+2\sqrt{\frac{1}{(1+{\tan }^2{\theta }_1)(1+{\tan }^2{\theta }_2)}}\le \frac{3}{4}.\stackrel{◯}{2}$$ From $\tan {\theta }_1\cdot \tan {\theta }_2=2$, we get $$\stackrel{◯}{2}\iff \frac{2+{\tan }^2{\theta }_1+{\tan }^2{\theta }_2}{5+{\tan }^2{\theta }_1+{\tan }^2{\theta }_2}+2\sqrt{\frac{1}{5+{\tan }^2{\theta }_1+{\tan }^2{\theta }_2}}\le \frac{3}{4}.\stackrel{◯}{3}$$ Write $x={\tan }^2{\theta }_1+{\tan }^2{\theta }_2$, then $$\stackrel{◯}{3}\iff 2\sqrt{\frac{1}{5+x}}\le \frac{14+x}{3(5+x)},$$ $$\text{that is,}36(5+x)\le 196+28x+x^2.\stackrel{◯}{4}$$ $$\text{Obviously,}\stackrel{◯}{4}\iff x-8x+16=(x-4{)}^2\ge 0.$$ $$\text{Hence, }\lambda =\frac{2\sqrt{3}}{3}.$$

When $n\ge 3$, there is no loss of generality in supposing ${\theta }_1\ge {\theta }_2\ge \cdots \ge {\theta }_n$, then $$\tan {\theta }_1\cdot \tan {\theta }_2\cdot \tan {\theta }_3\ge 2\sqrt{2}.$$ Since $\cos {\theta }_i=\sqrt{1-{\sin }^2{\theta }_i}<1-\frac{1}{2}{\sin }^2{\theta }_i$, so $$\cos {\theta }_2+\cos {\theta }_3<2-\frac{1}{2}({\sin }^2{\theta }_2+{\sin }^2{\theta }_3)<2-\sin {\theta }_2\cdot \sin {\theta }_3.$$ From ${\tan }^2{\theta }_1\ge \frac{8}{{\tan }^2{\theta }_2\cdot {\tan }^2{\theta }_3}$, we have $$\frac{1}{{\cos }^2{\theta }_1}\ge \frac{8+{\tan }^2{\theta }_2\cdot {\tan }^2{\theta }_3}{{\tan }^2{\theta }_2\cdot {\tan }^2{\theta }_3},$$ that is, $$\begin{array}{rl}\cos {\theta }_1& \le \frac{\tan {\theta }_2\cdot \tan {\theta }_3}{\sqrt{8+{\tan }^2{\theta }_2\cdot {\tan }^2{\theta }_3}}\\ & =\frac{\sin {\theta }_2\cdot \sin {\theta }_3}{\sqrt{8{\cos }^2{\theta }_2\cdot {\cos }^2{\theta }_3+{\sin }^2{\theta }_2\cdot {\sin }^2{\theta }_3}}.\end{array}$$ Hence $$\cos {\theta }_2+\cos {\theta }_3+\cos {\theta }_1$$ $$<2-\sin {\theta }_2\cdot \sin {\theta }_3\left[1-\frac{1}{\sqrt{8{\cos }^2{\theta }_2\cdot {\cos }^2{\theta }_3+{\sin }^2{\theta }_2\cdot {\sin }^2{\theta }_3}}\right].$$ Note that $$8{\cos }^2{\theta }_2\cdot {\cos }^2{\theta }_3+{\sin }^2{\theta }_2\cdot {\sin }^2{\theta }_3\ge 1$$ $$\begin{array}{rl}\iff & 8+{\tan }^2{\theta }_2{\tan }^2{\theta }_3\ge \frac{1}{{\cos }^2{\theta }_2\cdot {\cos }^2{\theta }_3}\\ & =(1+{\tan }^2{\theta }_2)(1+{\tan }^2{\theta }_3)\\ \iff & {\tan }^2{\theta }_2+{\tan }^2{\theta }_3\le 7.\end{array}$$ If this does not hold, then ${\tan }^2{\theta }_2+{\tan }^2{\theta }_3>7$, $${\tan }^2{\theta }_1\ge {\tan }^2{\theta }_2>\frac{7}{2}$$ so $$\cos {\theta }_1\le \cos {\theta }_2<\sqrt{\frac{1}{1+\frac{7}{2}}}=\frac{\sqrt{2}}{3}.$$ Thus $$\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3<\frac{2\sqrt{2}}{3}+1<2,$$ that is, the bound holds too.

$$\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3+\cdots +\cos {\theta }_n<n-1.$$ On the other hand, if we take ${\theta }_2={\theta }_3=\cdots ={\theta }_n=\alpha >0$, $\alpha \to 0$, then $${\theta }_1=\arctan \frac{2^n}{(\tan \alpha {)}^{n-1}}.$$ Obviously, ${\theta }_1\to \frac{\pi }{2}$, thus $$\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3+\cdots +\cos {\theta }_n\to n-1.$$ Consequently, we get $\lambda =n-1$.