## anonymous 4 years ago If $$\large a_1, a_2,a_3,\ldots,a_n$$ are real numbers, show that $\large\sum_{i=1}^{n}\sum_{j=1}^{n}i\cdot j\cdot \cos \left(a_i - a_j\right)\ge0$ @mukushla

1. anonymous

$\sum_{i=1}^{n} \sum_{j=1}^{n} ij \cos (a_{i}-a_{j}) = \sum_{i=1}^{n} \sum_{j=1}^{n} ij ( \cos a_{i} \cos a_{j} +\sin a_{i} \sin a_{j} )\\ =\sum_{i=1}^{n} i \cos a_{i} \sum_{j=1}^{n} j \cos a_{j}+\sum_{i=1}^{n} i \sin a_{i} \sum_{j=1}^{n} j \sin a_{j}\\=(\sum_{k=1}^{n} k \cos a_{k})^2+(\sum_{k=1}^{n} k \sin a_{k})^2\ge0$

2. anonymous

you're really good mukushla. thanks for helping me

3. anonymous

welcome :D