Problem
My attempt
From AM-GM inequality it can be shown that $$\sum_{i=1}^{n}(a_i+b_i)^2\geq4\sum_{i=1}^{n}a_ib_i$$ Therefore, we have: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}a_ib_i$$ Now, on expanding the RHS, we get: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\left({1\over a_1b_1}+{1\over a_2b_2}+.....{1\over a_nb_n}\right)\left(a_1b_1+a_2b_2+....a_nb_n\right)$$ $$=4(n+\left({a_2b_2\over a_1b_1}+{a_1b_1\over a_2b_2}+....{a_nb_n\over a_1b_1}+{a_1b_1\over a_nb_n}\right)+\left({a_3b_3\over a_2b_2}+{a_2b_2\over a_3b_3}+....{a_nb_n\over a_2b_2}+{a_2b_2\over a_nb_n}\right)+...+\left({a_nb_n\over a_{n-1}b_{n-1}}+{a_{n-1}b_{n-1}\over a_nb_n}\right))$$
Using the fact that $a+{1\over a}\geq2$ we get: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\left({1\over a_1b_1}+{1\over a_2b_2}+.....{1\over a_nb_n}\right)\left(a_1b_1+a_2b_2+....a_nb_n\right)\geq4(n+2(n-1+n-2+....+1))=4n^2$$
Is this proof correct? I am asking this because the proof given by the author of my textbook is quite different from this one.
Edit: Author's solution.