Prove that $n\sum_{i=1}^n a_ib_i \geq \sum_{i=1}^n a_i \cdot \sum_{i = 1}^n b_i.$

Question

Let $a_1 \leq a_2 \leq \cdots \leq a_n$ and $b_1 \leq b_2 \leq \cdots \leq b_n$, then prove that $$n\sum_{i=1}^n a_ib_i \geq \sum_{i=1}^n a_i \cdot \sum_{i = 1}^n b_i.$$

Attempt

The $n\displaystyle \sum_{i=1}^n a_ib_i$ makes me think of Cauchy-Schwarz but I am not sure how to use it. I could use AM-GM to say $n\displaystyle \sum_{i=1}^n a_ib_i \leq n\displaystyle \sum_{i=1}^n \dfrac{(a_i+b_i)^2}{4}$, but I am not sure how to relate that to the other terms.

Maybe need the restriction that the a's and b's are all nonnegative, or maybe all positive. — coffeemath, Jan 03 '16 at 04:28
You don't. Maybe that is a hint to use the rearrangement inequality? — user19405892, Jan 03 '16 at 04:29
@Winther Yes, but there may be a more clever way using inequalities. — user19405892, Jan 03 '16 at 04:32
Check https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality — Macavity, Jan 03 '16 at 04:36

score 3 · Answer 1 · answered Aug 26 '16 at 17:48

Observe that \begin{align*} \sum_{i=1}^n a_i \sum_{i=1}^n b_i - n \sum_{i=1}^n a_ib_i &=\sum_{i,j=1}^na_ib_j - n \sum_{i=1}^n a_ib_i\\ &=\frac{1}{2}\left[\sum_{i, j=1}^n(a_ib_j + a_j b_i) - \sum_{i, j=1}^n(a_ib_i +a_jb_j) \right]\\ &=\frac{1}{2}\sum_{i, j=1}^n(a_ib_j+a_jb_i-a_ib_i-a_jb_j)\\ &=-\frac{1}{2}\sum_{i, j=1}^n(a_j-a_i)(b_j-b_i)\\ &\le 0. \end{align*}

Prove that $n\sum_{i=1}^n a_ib_i \geq \sum_{i=1}^n a_i \cdot \sum_{i = 1}^n b_i.$

1 Answers1