Математикадан «Туймаада» олимпиадасы. Жоғары лига. 2003 жыл

$(0,\pi /2)$ интервалында берілген, кез-келген ${{\alpha }_{1}}$, ${{\alpha }_{2}}$, $\ldots$, ${{\alpha }_{n}}$ үшін, келесі теңсіздікті дәлелдеңіз: $\left( \dfrac{1}{\sin {{\alpha }_{1}}}+\dfrac{1}{\sin {{\alpha }_{2}}}+\ldots+\dfrac{1}{\sin {{\alpha }_{n}}} \right)\left( \dfrac{1}{\cos {{\alpha }_{1}}}+\dfrac{1}{\cos {{\alpha }_{2}}}+\ldots+\dfrac{1}{\cos {{\alpha }_{n}}} \right)\le 2{{\left( \dfrac{1}{\sin 2{{\alpha }_{1}}}+\dfrac{1}{\sin 2{{\alpha }_{2}}}+\ldots+\dfrac{1}{\sin 2{{\alpha }_{n}}} \right)}^{2}}$.

Комментарий/решение

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

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

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

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

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

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

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

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