# Prove that, $\sum_{i = 1}^n \frac{1}{a_ib_i} \sum_{i = 1}^n (a_i+b_i)^2 \geq 4n^2$

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.

• You can use AM-GM again... – Joey Zou Dec 31 '15 at 16:11
• So we have $\sum_{i,j} \dfrac{4a_ib_i}{a_jb_j} \geq \sum_{i,j} \dfrac{4a_ib_i}{\dfrac{(a_j+b_j)^2}{4}}$? – user19405892 Dec 31 '15 at 16:13
• The index in the first sum is $i$ or $j$? – Jimmy R. Dec 31 '15 at 16:13
• same index $i$ in two sums in Title is confusing. – p Groups Dec 31 '15 at 16:13
• i modified an i to a j to make it clearer – DanielWainfleet Dec 31 '15 at 16:38

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 }
• My first thought was that for a given pair $(a_i,b_i)$, if we keep $c_i=a_i+b_i$ constant, the LHS is minimized when $a_i=b_i$. This reduces the Q to $\sum_j(2/c_j)^2\cdot \sum_ic_i^2\geq 4n^2$. – DanielWainfleet Dec 31 '15 at 16:44