来个高手看看这道题

人间风雪客

期待你精彩的解答，脑子转不动了，唉

Ly-lie

这是一种证法（第一小问）
令$ x_i=\tan ^2\theta _i $,$ s=\sum_{i=1}^n\tan \theta _i $
$$ LHS=(\sum_{i=1}^n\tan \theta _i)(\sum_{i=1}^n\cos\theta _i)=(\sum_{i=1}^n\sec \theta _i-\sum_{i=1}^n\frac{\tan ^2\theta _i}{\sec \theta _i})\   $$\[ \leqslant (\sum_{i=1}^n\sec \theta _i-\frac{(\sum_{i=1}^n\tan \theta _i)^2}{\sum_{i=1}^n\sec \theta_i })(\sum_{i=1}^n\tan \theta _i)=(s^2-t^2)\frac{t}{s} \]
下证\[ (s^2-t^2)\frac{t}{s}\leqslant \frac{n^2}{\sqrt{n+1}} \]由柯西不等式\[ (n+1)(1+\tan ^2\theta_ i)=(n+1)(2\tan ^2\theta _i+\sum_{i\not=j}\tan ^2\theta _i)\geqslant (\sum_{i=1}^n\tan \theta _i+\tan \theta _i)^2 \]于是$$ \frac{1}{\cos \theta _i}\geqslant \frac{1}{\sqrt{n+1}}(\sum_{i=1}^n\tan \theta _i+\tan \theta _i),s\geqslant t\sqrt{n+1} $$因为$ st-\frac{t^3}{s} $关于s递增且\[s=\sum_{i=1}^n\sqrt{1+\tan ^2\theta _i}\leqslant [\sum_{i=1}^n(1+\tan ^2\theta _i)]^\frac{1}{2}\sqrt{n}=\sqrt{n(n+1)} \]故\[ (s^2-t^2)\frac{t}{s}\leqslant t\sqrt{n(n+1)}-\frac{t^3}{\sqrt{n(n+1)}} \]只需\[ n(n+1)t-t^3\leqslant n^2\sqrt{n} \]即\[ (t-\sqrt{n})(t^2+t\sqrt{n}-n^2)\geqslant 0 \]由于\[ t\leqslant \frac{s}{\sqrt{n+1}}\leqslant \sqrt{n} \]只需$ t^2+t\sqrt{n}-n^2\leqslant 0 $,这是因为\[t^2+t\sqrt{n}\leqslant n+n=2n\leqslant n^2 \]