An inequality problem - Solution explanation I have recently been reading about inequalities, and in this section discussing AM-GM inequalities, I was faced by this question that I couldn't understand its solution. Note: I cant find a relationship between the solution and AM-GM. I would be very thankful if someone can take me through the solution.
The problem is as follows :-
Let $a_1,a_2,...,a_n$ be positive numbers such that $\frac{1}{1+a_1}+\frac{1}{1+a_2}+...+\frac{1}{1+a_n} = 1$. Prove that
$$\sqrt{a_1}+...+\sqrt{a_n}\ge(n-1)(\frac{1}{\sqrt{a_1}}+...+\frac{1}{\sqrt{a_n}})$$
The solution was :-
$\sum_{i=1}^n \frac{1}{1+a_i} = 1 \Rightarrow \sum_{i=1}^n \frac{a_i}{1+a_i} = n-1$
Observe that
$$\sum_{i=1}^n \sqrt{a_i}-(n-1)\sum_{i=1}^n \frac{1}{\sqrt{a_i}}=\sum_{i=1}^n \frac{1}{1+a_i}\sum_{i=1}^n \sqrt{a_i} - \sum_{i=1}^n \frac{a_i}{1+a_i}\sum_{i=1}^n \frac{1}{\sqrt{a_i}}$$
$$= \sum_{i,j} \frac{a_i-a_j}{(1+a_j)\sqrt{a_i}}=\sum_{i\gt j} \frac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}}$$
Since $1 \ge \frac{1}{1+a_i}+\frac{1}{1+a_j}=\frac{2+a_i+a_j}{1+a_i+a_j+a_ia_j}$, we can deduce that $a_ia_j \ge 1$. Hence the terms of the last sum are positive.
I am really struggling to understand this proof and if someone could simplify it and (maybe show where it is related to AM-GM inequalities), I would be very grateful.
 A: The explanation is very dense and skips a lot of steps.
For the first implication, observe that $\sum_{i=1}^n \frac{a_i}{1+a_i} = \sum_{i=1}^n \frac{(1+a_i)-1}{1+a_i}=n-1$.
The first equality after "Observe that" is better written as:
$$
\left(\sum_{i=1}^n \sqrt{a_i}\right)-(n-1)\left(\sum_{i=1}^n \frac{1}{\sqrt{a_i}}\right)=
\left(\sum_{j=1}^n \frac{1}{1+a_j}\right)\left(\sum_{i=1}^n \frac{a_i}{\sqrt{a_i}}\right) - \left(\sum_{j=1}^n \frac{a_j}{1+a_j}\right)\left(\sum_{i=1}^n \frac{1}{\sqrt{a_i}}\right).
$$
The second equality follows from an algebraic simplification of the RHS.
The final equality is trickier. We break the double sum over $i,j$ into three pieces: (a) $j>i$, (b) $j=i$, and (c) $j<i$. The (b) terms drop out. The (a) terms can be written
$$\sum_i\sum_{j>i}\frac{a_i-a_j}{(1+a_j)\sqrt{a_i}}
\stackrel{(*)}=\sum_j\sum_{i>j}\frac{a_j-a_i}{(1+a_i)\sqrt{a_j}}
=-\sum_j\sum_{i>j}\frac{a_i-a_j}{(1+a_i)\sqrt{a_j}}
$$
where in $(*)$ we swap the index $i$ with the index $j$. Combining this with the (c) terms we get
$$\sum_{i>j}\left(\frac{a_i-a_j}{(1+a_j)\sqrt{a_i}}-\frac{a_i-a_j}{(1+a_i)\sqrt{a_j}}\right)
$$
which, after some algebra, is the same as the final expression. The final deduction "$a_ja_j\ge1$" follows from rearranging the inequality $1 \ge\frac{2+a_i+a_j}{1+a_i+a_j+a_ia_j}$. Since each of the $a$'s is positive, it follows that $\sqrt{a_i}\sqrt{a_j}=\sqrt{a_ia_j}\ge1$. We deduce that every term in the final summation is a product of non-negative items, and the final summation is non-negative. 
I don't see AM-GM anywhere in the proof.
