Олимпиада Туймаада по математике. Старшая лига. 2003 год


Докажите, что для любых $\alpha_1$, $\alpha_2$, $\dots$, $\alpha_n$, принадлежащих интервалу $(0,\pi/2)$, $$\left( {1\over \sin \alpha_1} + {1\over \sin \alpha_2} + \ldots + {1\over \sin \alpha_n} \right) \left( {1\over \cos \alpha_1} + {1\over \cos \alpha_2} + \ldots + {1\over \cos \alpha_n} \right) \leq $$ $$\leq 2 \left( {1\over \sin 2\alpha_1} + {1\over \sin 2\alpha_2} + \ldots + {1\over \sin 2\alpha_n} \right)^2 .$$
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Комментарий/решение:

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2019-06-10 13:06:54.0 #

$$ \Bigg( \sum_{1\leq k \leq n} \frac{1}{\sin{\alpha_k}}\Bigg)\Bigg( \sum_{1\leq k \leq n} \frac{1}{\cos{\alpha_k}}\Bigg)\leq 2 \Bigg( \sum_{1\leq k \leq n} \frac{1}{\sin{2 \alpha_k}}\Bigg)^2\qquad (1)$$

$$ (1) \Leftrightarrow 2 \Bigg( \sum_{1\leq k \leq n} \frac{1}{\sin{\alpha_k}}\Bigg)\Bigg( \sum_{1\leq k \leq n} \frac{1}{\cos{\alpha_k}}\Bigg)\leq \Bigg( \sum_{1\leq k \leq n} \frac{1}{\sin{\alpha_k}\cdot \cos{\alpha_k}}\Bigg)^2\qquad (2)$$

$$ \forall \alpha_k \in (0,\pi /2): 0<\sin{\alpha_k}<1 \Rightarrow \frac{1}{\sin{\alpha_k}}>1$$

$$ x_k= \frac{1}{\sin{\alpha_k}}, \quad y_k=\frac{1}{\cos{\alpha_k}} \Leftrightarrow x_ky_k=\sqrt{x_k^2+y_k^2}$$

$$ \forall x_k>1, \quad \forall y_k>1 \qquad 2\Bigg( \sum_{1\leq k \leq n} x_k\Bigg)\Bigg( \sum_{1\leq k \leq n}y_k\Bigg)\leq \Bigg( \sum_{1\leq k \leq n} \sqrt{x_k^2+y_k^2}\Bigg)^2\qquad (3)$$

$$ 2\Bigg( \sum_{1\leq k \leq n} x_k\Bigg)\Bigg( \sum_{1\leq k \leq n}y_k\Bigg)\leq\Bigg( \sum_{1\leq k \leq n} x_k\Bigg)^2+\Bigg( \sum_{1\leq k \leq n}y_k\Bigg)^2=$$

$$=\sum_{1\leq k \leq n}(x_k^2+y_k^2)+2\cdot\sum_{1\leq j\ne k \leq n}(\underbrace{x_jx_k+y_jy_k}_{\leq})\leq $$

$$\leq\sum_{1\leq k \leq n}(x_k^2+y_k^2)+2\cdot\sum_{1\leq j\ne k \leq n}(\sqrt{x_j^2+y_j^2}\cdot \sqrt{x_k^2+y_k^2})=\Bigg( \sum_{1\leq k \leq n} \sqrt{x_k^2+y_k^2}\Bigg)^2$$