Bariz - Problem

Source: Tuymaada 2003, day 2, problem 1.

Tags: trigonometry, inequalities proposed, inequalities

mathmanman

05.05.2007 15:18

Prove that for every $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ in the interval $(0,\pi/2)$ \[\left({1\over \sin \alpha_{1}}+{1\over \sin \alpha_{2}}+\ldots+{1\over \sin \alpha_{n}}\right) \left({1\over \cos \alpha_{1}}+{1\over \cos \alpha_{2}}+\ldots+{1\over \cos \alpha_{n}}\right) \leq\] \[\leq 2 \left({1\over \sin 2\alpha_{1}}+{1\over \sin 2\alpha_{2}}+\ldots+{1\over \sin 2\alpha_{n}}\right)^{2}.\] Proposed by A. Khrabrov

Let $a_{i}: = \sin \alpha_{i}$ and $b_{i}: = \cos \alpha_{i}$ then $a_{i},b_{i}\in (0,1)$ and we have to show \[\sum_{i=1}^{n}1/a_{i}\sum_{j=1}^{n}1/b_{j}\le 1/2 \Big(\sum_{i=1}^{n}1/(a_{i}b_{i}) \Big)^{2}, \] i.e. \[\sum_{i,j}1/(a_{i}b_{j}) \le \sum_{i,j}1/(2a_{i}a_{j}b_{i}b_{j}). \] Thus it suffices to show that for all $i,j$ we have \[1/(a_{i}b_{j})+1/(a_{j}b_{i}) \le 1/(a_{i}a_{j}b_{i}b_{j}) \quad \text{ i.e. }a_{i}b_{j}+a_{j}b_{i}\le 1. \] But $a_{i}b_{j}+a_{j}b_{i}= \sin(\alpha_{i}+\alpha_{j})$. So we are done.