How is the last equation derived in this equality?

Question

From Inequalities A Mathematical Olympiad Approach, page 122

Observe that, $$\sum_{i = 1}^n \sqrt{a_i}-(n-1)\sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \sum_{i = 1}^n \dfrac{1}{1+a_i}\sum \sqrt{a_i} - \sum_{i = 1}^n \dfrac{a_i}{1+a_i} \sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}}\\ = \sum_{i,j} \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} = \sum_{i>j} \dfrac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}}$$

I get that one $1+a_j$ factor in the denominator is from the given summation condition and the we applied $a^2-b^2$ for square root in the numerator to get two factors. But where do other factors come from? Also in the summation in the last step, how did we say $i>j$?

Answer 1

Note that $\sum_{i,j}$ means $\sum_{i=1}^n\sum_{j=1}^n$ and therefore \begin{align}\sum_{i=1}^n\sum_{j=1}^n \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} &=\sum_{i=1}^n\sum_{j=1}^{i-1} \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}}+ \sum_{j=1}^n\sum_{i=1}^{j-1} \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}}\\ &=\sum_{i=1}^n\sum_{j=1}^{i-1} \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}}+ \sum_{i=1}^n\sum_{j=1}^{i-1} \dfrac{a_j-a_i}{(1+a_i)\sqrt{a_j}}\\ &=\sum_{i=1}^n\sum_{j=1}^{i-1}\left(\dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}}+\dfrac{a_j-a_i}{(1+a_i)\sqrt{a_j}}\right)\\ &=\sum_{1\leq j<i\leq n}\dfrac{(a_i-a_j)((1+a_i)\sqrt{a_j}-(1+a_j)\sqrt{a_i})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}}\\ &=\sum_{1\leq j<i\leq n}\dfrac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} \end{align} where at the last step we applied $$(a_i-a_j)=(\sqrt{a_i}-\sqrt{a_j})(\sqrt{a_i}+\sqrt{a_j})$$ and $$((1+a_i)\sqrt{a_j}-(1+a_j)\sqrt{a_i})= (\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j}).$$