Let $a_1\le\dots\le a_n$ and $b_1\le\dots\le b_n$. For all permutation $\sigma\in S_n$ prove that: $$\sum_{i=1}^na_ib_{n-i+1}\le\sum_{i=1}^na_ib_{\sigma(i)}\leq\sum_{i=1}^na_ib_i.$$
Let $a_1\le\dots\le a_n$ and $b_1\le\dots\le b_n$. Prove that: $$n\sum_{i=1}^na_ib_{n-i+1}\le\sum_{i=1}^na_i\sum_{i=1}^nb_i$$ and $$\sum_{i=1}^na_i\sum_{i=1}^nb_i\leq n\sum_{i=1}^na_ib_i.$$