TITLE: Prove that, $\sum_{i = 1}^n \frac{1}{a_ib_i} \sum_{i = 1}^n (a_i+b_i)^2 \geq 4n^2$
QUESTION [5 upvotes]: Let $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ be positive numbers. Prove that, $$\displaystyle \sum_{j = 1}^n \dfrac{1}{a_jb_j} \sum_{i = 1}^n (a_i+b_i)^2 \geq 4n^2.$$

I was thinking of using AM-GM. We have $a_ib_i \leq \dfrac{(a_i+b_i)^2}{4}$. So we can say $\displaystyle \sum_{i,j} \dfrac{(a_i+b_i)^2}{a_jb_j} \geq \sum_{i,j} \dfrac{4a_ib_i}{a_jb_j}$ but I don't know what to do next.

REPLY [11 votes]: By the AM-GM inequality, $\sqrt{a_i/b_i}+\sqrt{b_i/a_i}\ge 2$. Combine this with Cauchy-Schwarz:
$$
\eqalign{
2n&=\sum_{i=1}^n 2\le\sum_{i=1}^n\left(\sqrt{a_i\over b_i}+\sqrt{b_i\over a_i}\right)\cr
&=\sum_{i=1}^n{a_i+b_i\over\sqrt{a_ib_i}}\cr
&\le\sqrt{\sum_{i=1}^n(a_i+b_i)^2\sum_{j=1}^n{1\over a_jb_j}}.\cr
}
$$